How do you evaluate ${log}_{2} (64) = x$ $Log_2 (64)=x$ ?

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Answer 1 · 2015-02-04T13:07:35

You can use the definition of logarithm:
${log}_{a} (b) = x$ $log_a(b)=x$ $\to$ $->$ $a^{x} = b$ $a^x=b$

In your case you get:

$2^{x} = 64$ $2^x=64$ which can be written as:

$2^{x} = 2^{6}$ $2^x=2^6$ (because $2^{6} = 64$ $2^6=64$ )

and finally (same base, $2$ $2$ and comparing the exponents):

$x = 6$ $x=6$

How do you evaluate log2(64)=xLog_2 (64)=x?