Question

(2^n+4)-(2^n+3)=3 solution this question

Answer 1

Answer:

$n=\frac{log\frac{3}{8}}{log2}$

Step-by-step explanation:

Given:

$2^{n+4}-2^{n+3}=3$

$2(2^{n+3})-2^{n+3}=3$

$2^{n+3}=3$

Take logarithm on both sides we get

$log2^{n+3}=log3$

$(n+3)log2=log3$

$n+3=\frac{log3}{log2}$

$n=\frac{log3}{log2}-3$

$n=\frac{log3-3log2}{log2}$

$n=\frac{log3-log2^3}{log2}$

$n=\frac{log3-log8}{log2}$

$n=\frac{log\frac{3}{8}}{log2}$