Solve-for-x-x-log-5-3x-12x-

Question Number 124880 by WilliamsErinfolami last updated on 06/Dec/20

S o l v e f o r x

x^{{l o g}_{5} 3 x} = 12 x

Answered by mr W last updated on 06/Dec/20

(\log_{5} 3 x) (\log_{5} x) = \log_{5} (12 x)

(\log_{5} 3 + \log_{5} x) (\log_{5} x) = \log_{5} 12 + \log_{5} x

(\log_{5} 3 + t) t = \log_{5} 12 + t

t^{2} - (1 - \log_{5} 3) t - \log_{5} 12 = 0

t = \frac{1 - \log_{5} 3 \pm \sqrt{{(1 - \log_{5} 3)}^{2} + {4 \log}_{5} 12}}{2}

\log_{5} x = \frac{1 - \log_{5} 3 \pm \sqrt{{(1 - \log_{5} 3)}^{2} + {4 \log}_{5} 12}}{2}

\Rightarrow x = 5^{\frac{1 - \log_{5} 3 \pm \sqrt{{(1 - \log_{5} 3)}^{2} + {4 \log}_{5} 12}}{2}}

= 9.693765, 0.171932

Answered by aleks041103 last updated on 06/Dec/20

{l o g}_{x} (12 x) = {l o g}_{5} (3 x)

1 + {l o g}_{x} (12) = {l o g}_{5} 3 + {l o g}_{5} x

1 + \frac{{l o g}_{5} 12}{{l o g}_{5} x} = {l o g}_{5} 3 + {l o g}_{5} x

t = {l o g}_{5} x \Rightarrow x = 5^{t}

t + {l o g}_{5} 12 = t^{2} + ({l o g}_{5} 3) t

\Rightarrow t^{2} - (1 - {l o g}_{5} 3) t - {l o g}_{5} 12 = 0

t_{1, 2} = \frac{1 - {l o g}_{5} 3 \pm \sqrt{{(1 - {l o g}_{5} 3)}^{2} + 4 {l o g}_{5} 12}}{2}

\Rightarrow x_{1, 2} = 5^{t_{1, 2}}