Be Careful How You Approach This | Cambridge Final Year Math Problem | Where To Start From.

[@pierrickroger4431]

Why not use the natural logarithm right at the first step ? I got to the same conclusion within 4 steps

[@ChavoMysterio]

3^(n+4)-4^(n+3)=0
3^(n+4)=4^(n+3)
81(3ⁿ)=64(4ⁿ)
(¾)ⁿ=64/81
(¾)ⁿ=(8/9)²
n=2[log_¾(8/9)] ❤

[@Bertin-q3y]

X=3 +(ln3)/(ln(3/4)

[@dukassis4906]

2. log (8/9) na base 3/4, para ser mais exato

[@anderson11884]

😊wow!! Nice!!

[@antonyqueen6512]

You got it wrong. It’s x=0.81884 not 0.8168
It too simple to use log straight away and
x= [4Ln(3)-3Ln(4)]/[Ln(4)-Ln(3)]

[@ManojkantSamal]

^=read as to the power
*=read as square root
As per question
3^(x+4) - 4^(x+3)=0
So,
3^(x+4)=4^(x+3)
(3^x). (3^4)=(4^x). (4^3)
(3^x)/(4^x) =(4^3)/(3^4)
(3/4)^x ={4^3/3^3}×(1/3)
(3/4)^x={(4/3)^3}×(1/3)
{(3/4)^x}/{(3/4)^-3}=(1/3)
(3/4)^(x+3)=1/3
Take the log
log{(3/4)^(x+3)}=log(1/3)
(X+3)×log(3/4)=log(1/3)
X+3=log(3/4)/log(1/3)
X=(-3)+{log(3/4)/log(1/3)}
X=(-3)+{(log3-log4)/(log1-log3)}
=(-3)+{(log3-log4)/(0-log3)
=(-3)+{(log3-log4)/(-log3)}
=(-3)+{(log3/-log3)+(log4/log3)}
=(-3-1)+(log4/log3)
=(-4)+(log4/log3)
=(-4)+{log4 of base 3)....May be

[@ralphtrucksess7008]

The path was right but the result is false:
you made a mistake in line 11:
(X+3)×log(3/4)=log(1/3) is not equivalent to X+3=log(3/4)/log(1/3)
but this does:
(X+3)×log(3/4)=log(1/3) X+3=log(1/3)/log(3/4)
and I prefer your way :)
No bases of 3/4 and no conversions from base 3/4 to base 10

[@HalilSer505]

👏

[@簡欽慧]

糊塗

[@gerardomr8231]

No, desde el inicio se pudo aplicar logaritmos y de una manera más fácil, directa y sin tantos pasos, se hubiera llegado al resultado !!

[@torstenbroeer1797]

This was the way I did it and I got the right answer x=0.8188… within two minutes.
Not 0.8168 as this idiot. And I call him an idiot because he is making calculations for nearly a1/4 hour that don't bring him any closer to the result, not because he pressed the wrong key at one instance.