How do you evaluate ${log}_{16} (\frac{1}{4})$ $log_16(1/4)$ ?

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Answer 1 · 2016-03-23T14:01:34

$x = - \frac{1}{2}$ $x=-1/2$

Let ${log}_{16} (\frac{1}{4}) = x$ $log_16(1/4)=x$ , then $16^{x} = \frac{1}{4}$ $16^x=1/4$ or

${(2^{4})}^{x} = \frac{1}{2^{2}}$ $(2^4)^x=1/(2^2)$ or

$2^{4 x} = 2^{- 2}$ $2^(4x)=2^(-2)$

Hence $4 x = - 2$ $4x=-2$ i.e. $x = - \frac{2}{4} = - \frac{1}{2}$ $x=-2/4=-1/2$

Answer 2 · 2016-03-23T14:10:13

Let's write it in exponential form.

${log}_{a} b = x \to a^{x} = b$ $log_ab = x -> a^x = b$

${log}_{16} (\frac{1}{4}) = x \to 16^{x} = \frac{1}{4}$ $log_16(1/4) = x -> 16^x = 1/4$

Solve for x by putting everything in the same base.

${(2^{4})}^{x} = \frac{1}{2^{2}}$ $(2^4)^x = 1/(2^2)$

$2^{4 x} = 2^{- 2}$ $2^(4x) = 2^(-2)$

$x = - \frac{1}{2}$ $x = -1/2$

Therefore, ${log}_{16} (\frac{1}{4}) = - \frac{1}{2}$ $log_16(1/4) = -1/2$

Practice exercises:

a). ${log}_{9} (\frac{1}{27})$ $log_9(1/27)$

b) ${log}_{x} (81) = 4$ $log_x(81)= 4$

c) ${log}_{2 x + 1} 11 x = 2$ $log_(2x + 1)11x = 2$

Good luck!

How do you evaluate log16(14)log_16(1/4)?