How do you simplify ${log}_{4} 8$ $log_4 8$ ?

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Answer 1 · 2016-05-07T16:10:59

Use the logarithmic properties:

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

${log}_{a} (b) = \frac{{log}_{c} (b)}{{log}_{c} (a)}$ $log_a(b)=log_c(b)/log_c(a)$

You can notice that $c = 2$ $c=2$ fits this case since $8$ $8$ can be derived as a power of $2$ $2$ . Answer is:

${log}_{4} 8 = 1.5$ $log_(4)8=1.5$

${log}_{4} 8$ $log_(4)8$

$\frac{{log}_{2} 8}{{log}_{2} 4}$ $log_(2)8/log_(2)4$

$\frac{{log}_{2} 2^{3}}{{log}_{2} 2^{2}}$ $log_(2)2^3/log_(2)2^2$

$\frac{3 \cdot {log}_{2} 2}{2 \cdot {log}_{2} 2}$ $(3*log_(2)2)/(2*log_(2)2)$

$\frac{3}{2}$ $3/2$

$1.5$ $1.5$

How do you simplify log48log_4 8?