How do you solve ${log}_{2} 32 = x$ $Log_2 32 = x$ ?

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Answer 1 · 2016-03-09T21:16:37

$x = 5$ $x=5$

The logarithmic expression can be exponentiated with a base of $2$ $2$ :

$2^{{log}_{2} 32} = 2^{x}$ $2^(log_2 32)=2^x$

The $2^{x}$ $2^x$ and ${log}_{2} x$ $log_2 x$ functions are inverses, which means that they undo one another, so we obtain the equation:

$32 = 2^{x}$ $32=2^x$

We can write $32$ $32$ as a power of $2$ $2$ :

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

Since the bases are equal, we know their powers are also equal, giving:

$x = 5$ $x=5$

Answer 2 · 2016-03-09T21:18:12

$x = 5$ $x=5$

Since $32 = 2^{5}$ $32=2^5$ , we see that

${log}_{2} 2^{5} = x$ $log_2 2^5=x$

Using the logarithm rule:

$log (a^{b}) = b \cdot log (a)$ $log(a^b)=b*log(a)$

Giving the equation:

$5 \cdot {log}_{2} 2 = x$ $5*log_2 2=x$

Since ${log}_{a} a = 1$ $log_a a=1$ ,

$x = 5$ $x=5$

How do you solve log232=xLog_2 32 = x?