How do you evaluate ${log}_{\sqrt{8}} 4096$ $log_sqrt8 4096$ ?

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Answer 1 · 2016-11-17T03:06:22

${log}_{\sqrt{8}} 4096 = 8$ $log_(sqrt8)4096=8$

Let ${log}_{\sqrt{8}} 4096 = x$ $log_(sqrt8)4096=x$

Using definition of logarithm that ${log}_{a} b = m$ $log_ab=m$ means $a^{m} = b$ $a^m=b$

${(\sqrt{8})}^{x} = 4096$ $(sqrt8)^x=4096$

expressing them in powers of $x$ $x$

or ${({\sqrt{2}}^{3})}^{x} = 2^{12}$ $(sqrt2^3)^x=2^12$

or ${({(2^{\frac{1}{2}})}^{3})}^{x} = 2^{12}$ $((2^(1/2))^3)^x=2^12$

or $2^{\frac{3 x}{2}} = 2^{12}$ $2^((3x)/2)=2^12$

or $\frac{3 x}{2} = 12$ $(3x)/2=12$

i.e. $x = 12 \times \frac{2}{3} = 8$ $x=12xx2/3=8$

Hence ${log}_{\sqrt{8}} 4096 = 8$ $log_(sqrt8)4096=8$

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