How to Solve Logarithmic Equations with Different Bases - The Change of Base Formula 

The lesson is well understood sir,thanks a lot.

Excellent tutorial covering, ahem, a lot of bases! No pun was intended!

I love this. One thing I wanted to add to gues video is that being log8(x) and log16(x) have base 2 in common, we could rewrite it as log2^3(x) and log2^4(x) and identifying a log base value would be the denominator when the change of base is applied, we could then rewrite the equation as 1/3log2(x)-1/4log2(x) and then proceed from there.

At starting point can be eliminate logx from boths side, then how can get x value?

I need some of your previous videos on logs

Thank you so much .

So far so good 👍

A man who makes totally time wasting short videos has 10 million subscribers in his channel but this channel is really awesome and has only approx 2 lakh. Not fair yaarrr

Can you say that cause X^3 = X^4 so it have to be 1 and 0 otherwise its not possible?

I used log.2 when using the change of base formula, this eliminated many of the steps. i only arrived at 1 solution, x = 1.

Really, exciting Vedio 🔥😇

00:00 It is even more simple when you use the formula: log.(a^m) (x) = (1/m) * log.a(x) Hence you get: log.8(x) - log.16(x) = 0 log.(2^3)(x) - log.(2^4)(x) = 0 (1/3) * log.2(x) - (1/4) * log.2(x) = 0 Now it is quite simple: [ (1/3) - (1/4) ] * log(x) = 0 [ (4/12) - (3/12) ] * log(x) = 0 (1/12) * log(x) = 0 || *12 log(x) = 0 log.10(x) = 0 x = 10^0 x = 1 "1" is the solution, because it is in the Domain which is D: x>0

Thanks a lot sir,

Sir, usually when doing this problem you also have to add two with log of x

I love this. Two logs have different base and are equal. How's that? So that x = 1. Just because no other points where different-base logs can meet. Telling this basic story during nine minutes - great.

There is a short solution. You solve the right side y= log.16 (x) by taken it to the base (2*8) . y is only a placeholder for the watched term. So (2*8)^y=x eq(1). This can be simplified to 2^y*8^y=x (eq2). Now the left side is taken to base 8. So y=log.8 (x) becomes 8^y=x eq(3) , which can be written as 2^y*2^y*2^y=x. Now taken the 3. root from that: 2^y=3.root(x) (eq4) eq(3) and eq(4) we set in eq(2) and we get 3.root(x)*x=x. We can see the possible solution are x=0 and x=1. But x=0 is no solution, because the original equation get infinity. It remains x=1

For those of you telling sir that he could have solved the problem in three steps keep your mouth shut, he's a professor he knows wat he's doing and his intentions was to clearly illustrate for students that might not readily know how to go about solving this problem problem, thank you.

Sir, you don't need to do that long since log_n(1)=0

Thanks for that

Change of base, seems to suggest Gauge equation: unit conversion transform method. Conductance, energy, conversion and energy transform ascertained. Thanks and have a nice day, thank you Sam.

The log function is not defined on the interval ]-∞, 0]. You MUST state that BEFORE begining any calculus, hence excluding 0 as a solution. Thus checking the answer x=0 as you do is unnecessary (not to say wrong).

i liked and watched the video sir

Sir, can you do this problem

Lo scopo dell'esercizio proposto e la sua soluzione, è anche di mostrare molte belle proprietà dei logaritmi e le molte ed eleganti vie del calcolo.

Hi Sir, just wonder about this part x^4 = x^3, Couldn't you just divide both sides by x^3 so you could find just the true answer right away? Thanks anyway way.

Sir please give solution for this If log a basex=log b base x thena=b if where a>0^b>0^ x>0 x is not equal to 1

u have made it an equation history.it could be solved more easierly

May God almighty bless you

At 1:34, you have Log X on both sides. If we divide by this we will have 1/Log 8 = 1/Log 16. How can this be?

I am understanding

What i did at the end where we have: X^⅓ = X^¼ is: Divide both sides by X^¼ X^⅓/X^¼=1 Move the X^¼ to the numerator and get: X^⅓*X^-¼=1 X^1/12=1 (X^1/12)^12=1^12 X=1

X should only be equal to 1 sir as log of zero to any base is under define

Only 1 fulfills the equation.

I watched and liked it

Dude! I did it the easier way out. First, what I did is, used the change of base formula ; then rewrote 16 as 2^4, and 8 as 2^3 ; then I used the power rule of logs "log(n^b) = b * log(n)" ; then I multiplied both of the fractions to get common denominators ; then I set just the numerator equal to zero after setting the whole fraction equal to 0 ; then I used the power rule of logs in reverse "b * log(n) = log(n^b)" ; then I used the quotient rule of logs ; then I used the quotient rule of exponents within the logs ; then I set log(x) to be equal to log(x) with base 10 ; then I converted to exponential form ; then noticed 10^0, which is equal to 1, and that is equal to x. What you are doing is the longer way out. you are checking for more possible solutions, but they are extraneous, so it would have been better if you did what I did no?

Why do you go long distance

This guy is an excellent teacher.

Instead of using base 10 it is easier if you use base 2 on the change of base

Sir the solution is maximum 3 steps only ....

At 2:59 log (x)/a = log(x)/b is impossible unless log (x) = 0 or x=1.

thanks

I didn't do it that same way but still had an answer. I got to a part of 4INX - 3INX which is equal to zero. Then I subtracted 3 from 4, which which made it e of x is equal to zero. Then I had x to be one.

03:45 You should rather put this "(1/4)*log(x)" form the right side to the left side (with the sign "+" changed to "-"): (1/3)*log(x) = (1/4)*log(x) (1/3)*log(x) - (1/4)*log(x) = 0 [ (1/3) - (1/4) ] * log(x) = 0 [ (4/12) - (3/12) ] * log(x) = 0 (1/12) * log(x) = 0 || *12 log(x) = 0 log.10(x) = 0 x = 10^0 x = 1 "1" is the solution, because it is in the Domain which is D: x>0

Fantastic explanation 👍

I am Bangladeshi 🥰🥰🥰 thanks a lot ❤️

Had a confusion about the base changing of a logarithm. It totally solved the problem. Appreciate it👏

Good

I almost solved this but got stuck at the end cause of splitting of x4-x3😅

Why the answer not x=1?

Where will be 12

C'est une farce?

I think x = {0,1,2,3,4,5,6,7} , what is your idea

Amazing..

Mr sanel he is a professor of what. I can also call you professor

These people were trying to outsmart the teacher.really? Why waste watching if you know already and know quick solution? ..this lecture is for those students or person who has difficulty and willing to understand it step by step. Just be humble guys, make ur own RU-vid channel..

I think we can solve this problem in another way avoiding all the complicated calculations Logically how can the power of 8 and 16 be same for the same value unless the value is 1 and power is 0 hence x has to be equal to 1

Can somebody solve 4^ = 2^x

Can somebody solve 4^x = 2^x

3^(x-1) =5

I got answer as 12

x=1

(log x)/(log 8) = (log x)/log16 Log 8 = log 16 8 = 16 Where did I make a mistake professor

I have a Little doubt, sir At 6:30, you got X⁴ = X³, so as the bases are equal, we get 4 = 3 but 4 ≠ 3, I am a bit confused, could you please reply However, Thank you very much for this amazing video, I have been actively watching a lot of your videos and they have really amazed me! Keep it up! hope to see more!

Log of x base2=0=>2^0=x=>x=1

It was obvious from the beginning that log x = 0, meaning x = 1, because the only way that raising x to two different powers gives an equality is if x = 1. This fellow used a sledgehammer to kill a fly. Far too complicated, tedious, and unnecessary.

It was obvious Why complicate things

log x/3log 2 =log x/4log 2, so log x=0, x=1😃

Ok

log(🐭) ×log(🐭) =[log(🐭)] ×2 or [log(🐭)]?😔😔

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forse 1

solo a vederla mi sembra non abbia soluzioni

Solution: (1) log8(x) - log16(x) = 0 According to logc(y) = loga(y)/loga(c) = same base (number/base), the following applies: log16(x) = log8(x)/log8(16) | applied in equation (1), results in: (1a) log8(x)-log8(x)/log8(16) = 0 |*log8(16)≠0 ⟹ (1b) log8(x)*log8(16)-log8(x) = 0 ⟹ (1c) log8(x)*[log8(16)-1] = 0 |/[log8(16)-1]≠0 ⟹ (1d) log8(x) = 0 ⟹ x = 8^0 = 1 Lösung: (1) log8(x)-log16(x) = 0 Nach logc(y) = loga(y)/loga(c) = gleiche Basis (Numerus/Basis) gilt: log16(x) = log8(x)/log8(16) | in Gleichung (1) angesetzt, ergibt: (1a) log8(x)-log8(x)/log8(16) = 0 |*log8(16)≠0 ⟹ (1b) log8(x)*log8(16)-log8(x) = 0 ⟹ (1c) log8(x)*[log8(16)-1] = 0 |/[log8(16)-1]≠0 ⟹ (1d) log8(x) = 0 ⟹ x = 8^0 = 1

Elementary school math

it long

mind twisted already

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Why is it so confusing😂

Pliz h

haaa

Too many steps.

Воды много, автор объём нагоняет...