How do you solve ${log}_{\frac{1}{4}} (x) + {log}_{2} (x^{2}) = 3$ $Log_(1/4) (x) + log _2 (x^2) =3$ ?

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Answer 1 · 2016-05-29T01:08:25

We must first put in a common base; this can be done by using the change of base rule ${log}_{a} (n) = \frac{log n}{log a}$ $log_a(n) = logn/loga$

$\frac{log x}{log (\frac{1}{4})} + \frac{{log x}^{2}}{log 2} = 3$ $(logx)/log(1/4) + logx^2/log2 = 3$

$\frac{log x}{log (2^{- 2})} + \frac{{log x}^{2}}{log 2} = 3$ $logx/log(2^-2) + logx^2/log2 = 3$

$\frac{log x}{- 2 log 2} + \frac{{log x}^{2}}{log 2} = 3$ $logx/(-2log2) + logx^2/log2 = 3$

$\frac{log x}{- 2 log 2} + \frac{- 2 {log x}^{2}}{- 2 log 2} = 3$ $logx/(-2log2) + (-2logx^2)/(-2log2) = 3$

${log}_{\frac{1}{4}} (x) + {log}_{\frac{1}{4}} (x^{- 4}) = 3$ $log_(1/4)(x) + log_(1/4)(x^-4) = 3$

${log}_{\frac{1}{4}} (x \times x^{- 4}) = 3$ $log_(1/4)(x xx x^-4) = 3$

$x^{- 3} = {(\frac{1}{4})}^{3}$ $x^-3 = (1/4)^3$

$\frac{1}{x^{3}} = \frac{1}{64}$ $1/x^3 = 1/64$

$x^{3} = 64$ $x^3 = 64$

$x = 4$ $x = 4$

Hopefully this helps!

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