WHat is x if $3 {log}_{2} (x + 1) = 15$ $3log_2 (x+1) = 15$ ?

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Answer 1 · 2015-11-10T15:14:20

$31$ $31$

$3 {log}_{2} (x + 1) = 15$ $3 log_2(x+1) = 15$

...divide both sides by 3

${log}_{2} (x + 1) = 5$ $log_2(x+1) = 5$

... apply $2^{a}$ $2^a$ on both sides in order to get rid of the ${log}_{2}$ $log_2$

$2^{{log}_{2} (x + 1)} = 2^{5}$ $2^(log_2(x+1)) = 2^5$

... use $2^{{log}_{2} (a)} = a$ $2^(log_2(a)) = a$ since the operations $2^{a}$ $2^a$ and ${log}_{2} (a)$ $log_2(a)$ are inverse

$x + 1 = 2^{5}$ $x + 1 = 2^5$

... take $1$ $1$ on both sides and simplify ....

$x = 2^{5} - 1 = 32 - 1 = 31$ $x = 2^5 - 1 = 32 - 1 = 31$

WHat is x if 3log2(x+1)=153log_2 (x+1) = 15?