How do you solve the equation ${log}_{b} 8 + 3 {log}_{b} n = 3 {log}_{b} (x - 1)$ $log_b8+3log_bn=3log_b(x-1)$ for n?

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Answer 1 · 2016-12-04T09:40:30

$n = \frac{x - 1}{2}$ $n=(x-1)/2$

${log}_{b} 8 + 3 {log}_{b} n = 3 {log}_{b} (x - 1)$ $log_b 8+3log_b n=3log_b(x-1)$

Now as $a {log}_{m} b = {log}_{m} b^{a}$ $alog_m b=log_m b^a$ and ${log}_{m} x + {log}_{m} y = {log}_{m} x y$ $log_m x+log_m y=log_m xy$ ,

we can simplify above as under.

${log}_{b} 8 + 3 {log}_{b} n = 3 {log}_{b} (x - 1)$ $log_b 8+3log_b n=3log_b(x-1)$

or ${log}_{b} 8 + {log}_{b} n^{3} = {log}_{b} {(x - 1)}^{3}$ $log_b 8+log_b n^3=log_b(x-1)^3$

or ${log}_{b} (8 n^{3}) = {log}_{b} {(x - 1)}^{3}$ $log_b (8n^3)=log_b(x-1)^3$

or $8 n^{3} = {(x - 1)}^{3}$ $8n^3=(x-1)^3$

and $n^{3} = \frac{{(x - 1)}^{3}}{2^{3}}$ $n^3=(x-1)^3/2^3$

i.e. $n = \frac{x - 1}{2}$ $n=(x-1)/2$

How do you solve the equation logb8+3logbn=3logb(x−1)log_b8+3log_bn=3log_b(x-1) for n?