Solving an exponential equation with different bases 

1 divided by 0 (a 3rd grade teacher & principal both got it wrong), Reddit r/NoStupidQuestions ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-WI_qPBQhJSM.html

Interesting, I never knew you could do it that way. Here are my steps: 1. Take the natural log on both sides, & got: (x) (In2) = (x+2) (In 5) 2. Distribute: xIn2 = xIn5 + 2In5 3. Subtract xIn5 on both sides: xIn2 - xIn5 = 2In5 4. Factor the x out: x (In2 - In5) = 2In5 5. Divide: x = 2In5 / In2 - In5. This works everytime for exponential equations with difference bases.

u can actually also just do log both side right away and have: x log 2= (x+2) log5 x log 2 = x log5 + 2log5 x log 2-x log5= 2log5 x(log2-log5)=2log5 x=2log5/(log2-log5)

First thought: 2 ^ any power is even and 5^any power is odd. Hmmm

I enjoy learning math from you early in the morning and late at night, you're cool

An important point for students is to note that the reason we can divide by 2^x or 5^x, is that none of these can be equal to zero. Generally, however, when we divide by something that contains an unknown variable, we have to make sure to state it is different than zero and before we give our final answer, we check the case that it is zero.

this has genuinely explained log rules better than my teachers in a few minutes so thank you so much

For the last question, answer in the same format: Step 1: 3^2x = 6^(x - 2) Step 2: 9^x = (6^x) / 36 Step 3: 36 = (6^x) / (9^x) Step 4: 36 = (2/3)^x Step 5: x = logbase2/3(36) Step 6: x = log(36) / log(2/3) Step 7: x = 2log(6) / (log(2) - log(3)) Step 8: x = (2log(2) + 2log(3)) / (log(2) - log(3))

For the question at the end I got 2(log6)/(log2-log3), which can interestingly be rewritten as 2*((log2+log3)/(log2-log3))

What’s interesting about this is that when you get to (2/5)^x=25. Intuitively, whatever number you choose for x will affect the fraction of 2/5. But the larger the number x is, the faster the 5 will grow in comparison to the 2. And you will never be able to find a positive value for x where 2^x/5^x=25 because the denominator will always be bigger. Then if we use exponent rules, we can realize that our x value will have to be negative. A negative x will swap the numerator and denominator and allow for the fraction to be equivalent, for some negative x value, to 25

I thank you for proposing these math problems, the solution to the second problem was -2log(6)/(log3-log2)

X=-8.838045... I think that is the solution

Ok but why do he have a pokeball doe

This is so amazingly that everyone shouldn't complain about solving exponential logarithm with diff bases,but to be thankful...Thank you sir for this assistance once again.

Equation: 3^(2*x)=6^(x-2) 1. ln both sides: 2*x*ln3=(x-2)*ln(2*3) 2. Develop and use ln properties: x*2*ln3=x*(ln2+ln3)-2*(ln2+ln3) 3. Isolate x on the left hand side: x*(2*ln3-ln2-ln3)=-2*(ln2+ln3) 4. Solve for x: x=2*(ln2+ln3)/(ln2-ln3) which can be written as x=2*ln(2*3)/ln(2/3), We see that x0 and the denominator is

Your passion and enthusiasm are contagious. I always loved mathematics and would have loved to study it in its purest form. However I ended up in its applications. Nevertheless, I thoroughly enjoy your videos. :-) Keep up the good work.

The coolest part Is always the cute Little Pokéball in his hand. It's such a nice touch!

I thought of 2^x as e^xln2 and 5^x+2 as e^(x+2)ln5 and arrived to the same answer! Thanks to your previous teaching of course

To solve the problem you shared at the end 3^2x = 6^(x-2) Log both sides and transfer the power to the left 2xlog3 = (x-2)log6 Expand 2xlog3 = xlog6 - 2log6 Group like terms on the same side 2xlog3 - xlog6 = - 2log6 Extract the factor from the left side x(2log3 - log6) = - 2log6 Find x x = -2log6 / {2log3 - log6)

Tell me if I am wrong with my solution: 2^x = 5^(x+2) log 2^x = log 5^(x+2) x.log2 = (x + 2).log5 x(log 2 - log 5) = 2.log5 x.log(2/5) = 25 And doing further calculation we can calculate x. Pls tell me if I am wrong here as it could help me develop my mathematical skills.🙂🙂

Just log both sides immediately with either a natural or common log and save yourself the trouble of a weird base. Distribute x+2 so you get xlog(5)+2log(5)=xlog(2) Just get the xlog’s to one side, factor out the x X(log(2)-log(5))=2log(5) Then divide by everything other than X and that’s your answer without some weird base

3²ˣ = 6ˣ⁻² Expand the right side. 3²ˣ = 6ˣ(6⁻²) Divide both sides by 6ˣ. 3²ˣ/6ˣ = 6⁻² Simplify. (3²/6)ˣ = 1/36 (9/6)ˣ = 1/36 1.5ˣ = 1/36 Take the log base 1.5 of both sides. log₁.₅(1.5ˣ) = log₁.₅(1/36) x = log₁.₅(1/36) Use the change of base formula. x = (log 1.5)/(log 1/36) Expand the denominator (and numerator if you kept 1.5 as a fraction). x = (log 1.5)/(log 1 - log 36) Simplify. x = - log 1.5 / log 36

Thank you so much! I was just looking for it!

answer for the next question : 3^2x = 6^(x-2) taking log on both sides : log(3^2x) = log(6^(x-2)) 2xlog(3) = (x-2)log(6) 2xlog(3) = xlog(6) - 2log(6) 2xlog(3) = xlog(3) + xlog(2) - 2log(6) x(log(3) - log(2)) = -2(log(6)) x(log(3) - log(2)) = -2(log(3) + log(2)) x =-2(log(3) + log(2))/(log(3) - log(2)) here log can be replaced with natural log. Hope its correct.

He is actually a good teacher

Interesting way of doing it.

I just use a general formula for problems in the form, a^gx+c = b^fx+d, where a doesn't equal b. The formula is x = log ([b^d]/[a^c]). (a^g)/(b^f) (a^g)/(b^f) is the base if the way I wrote it was confusing.

I found a formula which states ..log base 1/n of a = n log (a) ...is it correct ?

I apply log or ln on both sides first and then solve, but tysm for showing me a different method :)

-2log6/log3 -log2 is the answer for the last question

Thank you brother

The main way I do it (I don't know if this is the most efficient way) is log both sides of the equation, move the exponent to the front. The equation for this scenario currently would be x(log(2))=x+2(log(5)). Divide either log on both sides of the equation. (For this equation I did log(5)/log(2)). After this you are left with x=x+2(log(5)/log(2)) or x=(roughly)2.32192x+4.64385. Subtract x from both side which leaves you with 0=1.32192x+4.64385. Subtract 4.64385 from both sides then divide by 1.32192. This gives the answer -3.512958, its not exact, (off by around 1/10,000 but it would be exact if you either use the ans key or write out the entire decimal.

Is anyone calculate by changing any number n to (e^ln(n)) first?

I did this instead... 3^2x = 6^(x-2) Taking log of each side 2x log(3) = (x-2) log(6) Subtract x log (6) from both sides 2x log(3) - x log(6) = -2 log(6) Take advantage of 2 log(3) = log(9) and -2 log(6) = log(1/36) To clean things up a bit x log(9) - x log(6) = log(1/36) Combine log(9) - log(6) to give log(9/6), which is log(3/2) x log(3/2) = log(1/36) Divide both sides by log(3/2) x = log(1/36) / log(3/2) Apply quotient rule to all logs, and multiply numerator and demoninator by -1. x = log(36) / (log(2) - log(3))

Despite learning this multiple times throughout my lifetime, I never retain any memory from my math classes

Before we start, is a PokeBall required to solve the equation?

Hello, I wanted to kindly remind you that I sent you a message regarding your videos. Do we have your permission to use them? We would be happy to know if there is any special requirement for publishing your videos that we could fulfill. Thank you! :) Sincerely, Divya

that was very helpful thx

I have been your subscriber for over 3 years now. You inspired me to start my own channel

Log_6/9(36) for the end question

I did it a different way and didn't need a calculator to solve it. I converted both sides to log10 and substituted the bases. 2=10^.3 and 5=10^.7 That left a simple algebra problem .3x=.7x+1.4. This simplifies down to x=-3.5.

In last we can put values of log on base 10 which is log5= 0.698, log2= 0.3010, log3=0.4771 🇮🇳

I'm in 8th grade and I solved his question at the end. 3^(2x) = 6^(x-2) (Pls pin me @just algebra) x = ~ - 8.838 How I solved it... I first did 3^(2x) = 6^x/6^2 Then I took Log (base 3) on both sides Log (base 3) 3^(2x) = Log (base 3)6^x/6^2 Then using logarithmic rules I got the right side = 2x because Log (base 3) of 3 = 1 and the exponent was 2x so it was just 2x On the left side, I got (using other logarithmic rules not going to type) x * Log (base 3) of 6 - 2 * Log (base 3) of 6 then Log (base 3) of 6 was in common so I factored that out on the left side So then I got: 2x = (x-2)*(Log (base 3) of 6) Then I divided by Log (base 3) of 6 on both sides to get: 2x/Log (base 3) of 6 = x-2 Then I divided by x both sides to get: 2/Log (base 3) of 6 = (x-2)/x Then I split up (x-2)/x to 1 - 2/x Then I added 2/x to both sides So I got 2/Log (base 3) of 6 + 2/x = 1 Then I subtracted 2/Log (base 3) of 6 on both sides getting: 2/x = 1 - 2/Log (base 3) of 6 Then I multiplied by x and divided by 1 - 2/Log (base 3) of 6 and I finally got X = 2/(1 - 2/Log (base 3) of 6) and then I used a calculator for the first time and I got X = ~ - 8.838. Please let me know if you understood my work or if you have any questions and also keep in mind I'm only 14 and a half so pls don't hate if I made a mistake Thanks

Good video for high school math!🤩🤩

I got -(ln 36)/(ln 1.5) for the last question👍

Thank you for the nice lesson

Just do it like this for unlimited values of x xlog2 = xlog5 + 2log5 2log5 = x(log2-log5) *"x = 2log5/(log2-log5)"* Now put any base Keep getting ∞ values of x

I'm a new subscriber. Just one question, do you have a video abt logarithms? If you don't have it, can you make a video abt it? I haven't seen that before.

Hello sir... I've not understood how you have cancelled the log 5 with denominator of log2-log5.. whereas we have negative sign before it... please elaborate

Could of used log from the very beginning to bring the exponents down.

For the assignment question, the answer is x= 2 log 6/(log 2 - log 3)

2ˣ = 5ˣ⁺ ² 2ˣ = 5ˣ • 5² 2ˣ / 5ˣ = 25 (2/5)ˣ = 25 log(2/5)ˣ = log 25 x log(2/5) = log 25 x = log 25 / log(2/5) x ≈ -3.513

By simple algebra are they equal lhs and rhs. If not do you think you can make them equal by using log jugglers?

(2/5)^x = 2^2 Taking log to the base 2 one gets x* ( 1 - lg(5) ) = 2 or x = 2/( 1 - lg(5) )

in the old days before slide-rules even, we had to use log tables, so I know that log(2) =0.3010 and so log(5)=0.6990 so taking logs 0.301x = (x+2)(0.6990) 0 = 0.398x + 1.3998 x=-1.3998/0.398 x= -3.5170

x=2(ln5/ln2)/(1-(ln5/ln2)) first equation

idk but i am feeling proud after solving it myself

I always used natural log because Ln is shorter then log and doing that over a half page of more complex problems saves time

Tks sit for this nice equation.Now here what would prompt you to apply logs while we normally use normal calculations

2^x=5^(x+2) Applying log(natural log) on both sides x•log2=(x+2)•log5 x/x+2=log5/log2 x/x+2=log base2(5) x=x•log base 2(5)+logbase2(25) x•{logbase2(2)- logbase2(5)}=logbase2(25) x=logbase2(25)/logbase2(2/5) x=logbase2/5(25) solved This question is solved using property logbasea(b)=logbase c(b)/logbase c(a)

xlog2 = (x+2) log5 (Base is 10 ) xlog2 = xlog5 + 2log5 x= 2log5/ log2 -log5 = -3.54

An exponent divided by an exponent doesn’t cancel the exponent? 4/25=625?

For the bonus question I got the following answer: x = Log[base 6/9](36) or x = 2*log(6)/(log(6) - 2 * log(3))

I just multiplie the power of the base 5 by log2(5) then let the exponent of LHS = the exponent of RHS

For the last one x= -log(36)/(log(3)-log(2))

The final answer is x= -3.5

The answer to question given by you at end is x = log2/3(36)

Is there a way to solve 2^X = 57^X+1? How about Squrare root X= X^3 solve for X?

I am new to this channel Content is great One question why does he have pokenon in his hand?

I got x = -8,84 ( approximately) Is that the right answer?

I did by integrating both sides wrt dx

1/(1 - ln(x)) = e^(1 - 1/x) how to solve for x?

I love your videos. Thank you.

Simply take log of both sides first and then solve. Much easier. X log 2 equals (x+2) log 5

Excellent sir..

I chose to divide both sides by 2^x (instead of 5^x) and my result is different: x = log5/2[1/25] = -3.51294159473206 ...I plugged it into that original equation and it checks out 🤪

I'd just take the log of the first equation, after which it's routine manipulation. Just fun I'll take the natural log.. 2^× = 5^(× + 2) becomes x* ln(2) = (x + 2) * ln(5) or x*(ln(2) - ln(5)) = 2 * ln(5) or x = 2 * ln(5)/(ln(2) - ln(5)) = 2/(ln(2)/ln(5) - 1) = -3.5129415947320600588642107219209 You could use log of any base instead of e (natural log, ln) to do this. In particular you could use log base 10 (aka "log" on your calculator). A good question is to ask why the base of logs doesn't matter. Answer: To convert bases consider definitions x = a ^ (log_a(x)) = b ^ (log_b(x) So taking log base a of both sides we get the base conversion formula log_a(x) = log_b(x) * log_a(b) Now let's use this to consider log_a(x)/log_a(y) = (log_b(x) * log_a(b))/(log_b(y)*log_a(b)) = log_b(x)/log_b(y) Note the cancelation of log_a(b). that is, for a ratio of logarithms to the same base, the value of the base doesn't matter. QED

Keeping my knowledge fresh as always.

Thank you

Just ln() both sides and solve for x. xln(2)=(x+2)ln5 x(ln(2)-ln(5))= 2ln(5). x=(2ln(5))/(ln(2]|ln(5)). Seems easier.

What is that red and white ball in his left hand? What is the purpose of it?

so helpful

Thank the heavens for numerical methods. There is no way I could have solved this. 😆

Thanks. That was fun. It has been a long time since I used log functions. Forgot how, LOL!

I still haven't learnt what log is yet or any of this in school but it looks fun so I'll learn it early

Thank you. This was excellent.

feeling proud that I actually solved it with my eyes and just came to prove that i’m right.

heres my method and its a little wierd 3^2x=6^x-2 3^2x=3^log3(6)(x-2) same bases so equate powers 2x=log3(6)(x-2) 2x=xlog3(6)-2log3(6) 2x-xlog3(6)=-2log3(6) x(2-log3(6))=-2log3(6) x=-2log3(6)/2-log3(6) x=-8.838

Gotta Solve ‘Em All

edit: I'm wrong but leaving my attempt anyway. answer before watching: 2 and 5 are prime. so any number that can be factored using one cannot be factored using only the other since neither can be factored into the other except 1 (which is 1 * x^0). This is the same as saying 2^x is never divisible 5 since 5 isn't divisible 2 and vice versa. So 2^x only equals 5^x when x is zero. this means 2^x and 5^(x+2) must equal 2^0 and 5^0. Since their powers are different, this problem is impossible. proof: 2^x = 2^0 x = 0 and 5^(x+2) = 5^0 x+2 = 0 sub in each other for zero x = x+2 0 = 2

We can also take the ln on both sides.

In last question, x = [2log6/(log6 - log9)]

2log6 /log 2 -log 3 is it right or wrong ..

for the question at the end: x=log3/6(6^(-2))

x=log25 with base: 2/5

U could use log10 at the very first step it could be easy

Well as i remember in grade 8 or 9, we used to have the same things but different number, but we didn't use log equation Just use the computer to calculate (2/5)^x = 25 Then we gonna have x

I just taught this to my summer school algebra 2 class.

If you write log without any base then by default the base is 10. So, how does this work?

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