6 to the (3x + 5) = 1, many don’t know where to start 

Yes, I saw this in a couple of seconds and it looks quite ridiculous to use logarithms here, except if the point was just training logarithms.

Well, Einstein said it well: "Always describe a problem as simply as possible, but no simpler." Same for programming, don't reinvent the wheel--unless you discover a better wheel."

@@BluesChoker01do you have any idea on where to start with programming…I tried to instal the python app in my laptop but it declined now I don’t know what to do from here ..

Yes the simplest and logically correct।

That's also how I solved it. But I guess he wants to teach the general method.

Yes, I think this a very roundabout way of doing this. He should remember that math tests have time limits.

@@paullambert8701 Definitely the quickest way when we know n^0 = 1, but if it is not =1 how do you solve?

@@johnwythe1409 :Yes, this only works if YadaYada =1. If not, then :(((

Well aren't you just the smartest boy?

@@paullambert8701 If you use logarithms it is also fairly quick. Reason it took so long in the video is because he spent several minutes teaching what logarithms are and how to use them.

Thank you!

How I did it too.

Your fast take was wrong. You put x=-3/5; the correct answer was x=-5/3. So the moral to the story is to slow down.

Precisely,,

ERROR FINAL!

Even less work than my solution - good job!

@@andrewhines1054 Greetings. Blessings.

That’s how I did it.

Good, correct, and detailed answer, Devon. I am a down, dirty, and done type of guy: (in my head) "Anything raised to the 0 power equals 1, so 3x+5 = 0, x=-5/3, next?) If Math Man John wanted to illustrate how to manipulate exponents and logarithms to solve the more generalized problems as he does in his explanation, he should have used a different problem. By using "1" here, he invites the analysis you and I did, but if it had read "... = 28", the "exponent is 0" trick wouldn't be useful.

👍

But the Earth isn't a perfect sphere. There'll be errors in navigation that needs adjustments to follow the appropriate path to the next port. Otherwise, you could encounter something unexpected and bad.

@@oahuhawaii2141 Which is what they do. Since it's very close the spherical trig gets you quite close to where you need to be and then you can make the adjustments that is part of the navigation process.

If you need to navigate at sea, you still need to be able to find your location the old way. Your GPS may be broken or fell overboard.

Anytime you see a more complicated equation or expression in an exponent try using logs to simplify it.

@@WitchidWitchid Indeed! :)

I took apart my 1970s Texas Instruments calculator and activated the square root button.

My sister got a TI SR10. It has built-in [x²], [√x], and [1/x] functions, but no parentheses or memory store/recall. It can display the result in mantissa with exponent format. Neatest thing since sliced bread for engineers and scientists.

For that equation to be true, 3x+5 must equal 0, so x=-5/3

Greetings. I can remember those days very well.

@@devonwilson5776 I also remember using a slide rule and it cost a whole lot less than even a basic calculator.

I was a poor graduate student armed only with a slide rule when scientific calculators first became available. I was the only student not using a scientific calculator because I also had a wife and kids to feed. Scientific calculators should never be banned from any exam when they are not banned from the job. Instead, the student should be required to justify every step taken to arrive at any answer obtained from a scientific calculator. For example, suppose my boss handed me this new equation he just discovered, (2.3)^(2x)-4*(2.3)^x+4=0, and ordered me to have a value for x by tomorrow. And by the way, scientific calculators are not allowed. I know several ways to solve this problem, but I can't think of a useful way to get a value for x without a scientific calculator. For example, substitute y=(2.3)^x into the given equation to get y^2-4y+4=0, which has roots y=[4±√(16-16)]/2=2, so (2.3)^x=2. Of course I could then take the natural log of this equation to get x*ln(2.3)=ln(2) and solve for x=ln(2)/ln(2.3), but without a scientific calculator I couldn't provide a numerical value for x without consulting my old "CRC Standard Mathematical Tables and Formulae" book.

If you can't trust those calculators, then they're useless. What you couldn't trust is the user, if he/she didn't understand how the device operates and its limitations. I think the real problems are that most early ones had no support for extended precision, floating point (mantissa & exponent), precedence rules, parentheses, and memory store/recall. They also lacked many common functions, such as roots, powers, exponentiation, logarithms, circular & hyperbolic functions along with their inverses. That means you still need to make sure you don't exceed the range of the calculator, use look-up tables for the common functions, be wary of precision errors introduced, and have scratch paper on hand to record intermediate results that will be hand-entered later on. That's where errors get introduced into the calculations. My sister got a TI SR10. It has built-in [x²], [√x], and [1/x] functions, but no parentheses or memory store/recall. It can display the result in mantissa with exponent format. Neatest thing since sliced bread for engineers and scientists.

Views is what count$

A friend got a Sinclair calculator. It was fun to play with, but we kept draining the batteries. He also got the computer, too.

My sister got a TI SR10, and later my brother got the HP 34C. I got a TI 35, TI 57, and HP 41C. Another friend got a TI 58. Another student got the TI 59. They were fun to use.

I still like slide rules.

you funny man, i like u

You’re right, when and where it applies. For example, x^3 = 27, which gives x = 3 after taking the cube root on both sides of that equation. Or, the same thing for x^2 = 100, say, which, taking the square root on both sides gives x = +/-10, apart from dimensions of time, length, etc., since there’s no such thing as negative length and time, etc. in which cases only the positive answer and not the negative answer is possible. Or, and probably more to your question and example but the very same thing nonetheless, and simply the same thing the other way around but using the inverse of roots we could have the square root of x = 10 where we’d then square both sides of the equation (the inverse of taking the square root on both sides) to get x = 100! Or we could have the cubed root of x = 3, where we’d simply raise the expressions on both sides of the equation to the third power (cube both sides of the equation, sort of speak) to get x = 27. Well, here, the same thing’s going on since logarithms and exponentials are inverses of each other, like exponents and roots are, one to the other, just as addition is to subtraction and division is to multiplication, and vice versa, etc. That is, we can use the inverse of a mathematical operation that’s in the form of an equation to solve that equation, or to check for the correctness of it. In other words, we check our answers to our addition for their correctness by its inverse operation of subtraction, and vice versa. Same with multiplication and division with respect to each other, etc.

We used slide rules when I was a junior, but we could use calculators as seniors. The school bought them and we could borrow one. The first Calc I used was the HP which used polish infix notation. It did everything but graph. Switched to TI as undergrad. Returned 20 years later for advanced degree, and we used a nice--TI graphing calculator. But for advanced Calc, discrete math, etc--we used laptops and Maple and Mathematica (now Wolfram). Once mastering these tools, you just couldn't miss homework problems. The best thing about these tools, was the ability to try all sorts of approaches. If an approach failed, you learned something. Playing around and thinking really expands your depth of knowledge. One thing to remember is there are no timed tests in the workforce-- only deliverable dates. Sure, some things needed quickly reviewing, but some problems took months, so they were always in the background, to be attacked as insight lit up connected neurons :-) Why is it that most "aha" moments occur in the shower, working out or in dreams?? 😮 I think the subconscious is handling a lot of work so the conscious mind can function on the immediate tasks at hand. Man, I got pretty metaphysical here.

Thanks for providing the explanation. It is helpful. I am arriving at a different answer than yours however. Could you take a look to see if you agree? In the third line of your calculation you are showing 6^(3x+4)=6^6. I am asking if this should instead be 6^(3x+4)=6^2. Thus 3x+4=2, 3x=-2, and x=-2/3.

Yes, but in his 2 problems you can do them easily, right? 6^(3*x+5) = 1 Take log base 6 of both sides: 3*x+5 = 0 x = -5/3 ≈ -1.666667 4^y = 10 y*log(4) = log(10) y = 1/2/log(2) We remember log(2) ≈ 0.30103, so y ≈ 1/0.60206 ≈ 1.660964 .

All good thing comes in three's

Greetings. I remember those days, for me 70's, in primary school, absolutely no calculator allowed. The good old days.

That’s because calculators are for students who understand and therefore know how to do the mathematical calculations without using a calculator. What if the calculator broke down and we didn’t know how to arrive at the answer way the instructor is doing here? It’s easy to hit 5 and then log10 on the calculator. But what if the calculator pegs out and break right at that moment and we didn’t know how to do the problem by hand because we didn’t know and understand how to do the problem by head and hand? Calculators are excellent for doing tests and even during class, but only after a subject has been thoroughly understood and learned by the head and done by the hand repetitiously until committed to both memory and understanding. Does no good to have students hit and peck 2 + 2 on the calculator when they don’t understand and know how to do addition through understanding and repetitive practice. What happens if the instructor gives a set of problems to be done within a reasonable time frame but says, “absolutely no calculators!” Then what! Calculators are for those who understand and know the work, they’re not for the uninitiated! The uninitiated will never understand and learn the subject matter properly and thoroughly by simply punching in log100 with a calculator… they need to first understand and learn why the calculator gives the answer 2! For example, why would I add 0.3456889731237 + 35.903578134556 = x by head and hand instead of using a calculator when I already understand and know how to add decimals by head and hand? It would consume way to much time and be completely insane for the teacher to have me to write that problem out by head and hand instead of using a calculator when the teacher knows that I’ve proven and therefore understand and know how to do the problem by head and hand! But for the uninitiated student they must necessarily begin by first understanding and knowing and doing the problem of adding decimals before they move on to calculators. Calculators are horrible for students who don’t first understand know how to do math by head and hand through practice, practice, practice! So… what do I do with 30.4567893400586 - 7.8967855422106 if I didn’t know how to subtract decimals by head and hand but knew how to peck out the answer on a calculator but all the calculators in the world were broke….?

No. It's -5/3. You forgot to divide 3x by 3.

You can also take the log instead of ln. Either way, you get the same answer

Using log to base 10 ( the normal log button): log 10 = 1 log 4 = 0.6 approx gives 1.66 when divided Using natural log (the ln button): log 10 = 2.3 approx log 4 = 1.386 approx again gives 1.66 I've no idea how you arrived at 1.22 (even tried mixing the two log functions). edit - I used your 1.22 and swapped around to find either: log 10/log 6.6 = 1.22 or log 5.426/log 4 = 1.22 Doesn't help much! On the Windows calculator (scientific mode) Simply input 10 press log - then divide by 4 press log i.e. (10 log)/(4 log) Not sure what calculator you are using.

…may depend on the hierarchical structure (program) built into your calculator. On mine I hit 10 (the 1 and 0 buttons), then hit the log 10 button, then hit the divide button, then hit the 4 button, then hit the log10 button, then hit the = button and the calculator gives the approximation 1.66……….. how you enter those terms in the proper sequence and correctly on a calculator depends on the calculator’s hierarchy structure.

Calculators aren’t all programmed the same. Some accept entries one way while others do it differently. Find out what type of hierarchical program your calculator has, or look at its usage instructions and you’ll get the correct answer.

He is using common logs, base 10.

Makes no difference.

I can vouch for this. When I went to Lake Stevens HS, the geometry teacher was a total bore. I didn't learn much and I had a hard time paying attention, so I would up with a D grade. When I transferred over to Mariner HS, I got into a trigonometry class. The teacher there was good, and I got a B. The teacher does make a huge difference, but the door swings both ways.

If you want to go for the infinite solution route with complex numbers, then you need to consider the polar form for 1, which is 1*e^(i*2*π*k) for all integer k: 6^(3*x+5) = 1 = 1*e^(i*2*π*k) (3*x+5)*ln(6) = 0 + i*2*π*k x = -5/3 + i*k*(π*2/3/ln(6)), for all integer k Note that for k = 0, x = -5/3 .

What happend to the 1 when you moved to step 3?

3x+5 = 0, 3x=-5, x=-5/3.

Especially easy when it's equal to one & only done so when it's equal to one. You can also do logs on both sides & doing a log of 1 is equal to zero & a log base 6 of 6^z gets you z = 0 |z=3x+5 & u get the same thing

@Mike-lx9qn Yes, but it seems unnecessarily complex and does not teach the principle that when bases are equal, then the exponents must be equal.

@@jamesharmon4994 "bases are equal, then the exponents must be equal." However, 5^0 = 1, 6^0 also = 1, so it is not clear that one must choose 6^0. Instead it is probably sufficient to know that any number to the 0 power is 1, thus the exponent must be equal to zero and it isn't really necessary to put the same base on both sides of the equal.

@thomasmaughan4798 True, but if the instructor requires "Show your work", it is.

Nope. You are way off.

If you want to go for the infinite solution route with complex numbers, then you need to consider the polar form for 1, which is 1*e^(i*2*π*k) for all integer k: 6^(3*x+5) = 1 = 1*e^(i*2*π*k) (3*x+5)*ln(6) = 0 + i*2*π*k x = -5/3 + i*k*(π*2/3/ln(6)), for all integer k Note that for k = 0, x = -5/3 .

This was how I solved this in 10 seconds.

First, If, as you say, when the exponent is 0 we have no numbers to multiply with, how then did or would we get or assume we’d get 0? Wouldn’t the answer as well as the question then be meaningless, if not absurd? And why do you say an exponent needs to be a positive real number? -3 is a negative real number and therefore less than 0 and bases have negative exponents such as -3. So are you saying that the base x in the expression x^(-3) doesn’t exist because negative 3 is negative and not positive and less than 0? Are you saying that we can’t have a negative exponent such x^(-3)! What about x/xxxx = 1/xxx = 1/x^3 = x^(-3)? That’s the same negative exponent, isn’t it, and it’s base exists! If the rule of division for exponents states that x^a/x^b = x^(a - b), then If a = b and we have x^a/x^b, then x^a/x^b = x^a/x^a, since a = b, = 1 =, by the rule, x^(a - a) = x^0 = 1. The proof (x^n/x^n = 1 = x^(n - n) = x^0 = 1) justifies the rule. Which is the same thing as x^a/x^b = x^(a - b) x^2/x^3 = x^(2 - 3) = x^(-1) = xx/xxx = 1/x = x^(-3). So that, where (a) equals (b), for example, both (a) and (b) equal 2, then, x^2/x^2 = xx/xx = 1 = x^(2 - 2) = x^0 = 1. You’re simply subtracting the power of the denominator from the power of the numerator. Which simply says that when the two powers (exponents) and their signs are equal and the same the quotient has to equal 1. By implication the rule is simply a logical and true conclusion, is all.