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

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

😊wow!! Nice!!

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

糊塗

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)] ❤

👏

^=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

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