Question

1/log10 base x +2 = 2/log 10 base 5 find the value of x​

Answer 1

Given:

1/log10 base x +2 = 2/log 10 base 5

To find:

Find the value of x​

Solution:

From given, we have,

1/log10 base x + 2 = 2/log 10 base 5

let log10 base x = u, so we have,

1/u + 2 = 2/log 10 base 5

u = - log 10 base 5/2(log 10 base 5 - 1)

again back substituting the value of u, we get,

log10 base x = - log 10 base 5/2(log 10 base 5 - 1)

⇒ 1/logx base 10 = - log 10 base 5/2(log 10 base 5 - 1)

2(log 10 base 5 - 1) = - log 10 base 5 × logx base 10

2(log 10 base 5 - 1) = - ln x / ln 5

2(log 10 base 5 - 1) ln 5 = - ln x

ln x = - 2(log 10 base 5 - 1) ln 5

x = e^ {- 2(log 10 base 5 - 1) ln 5}

x = [ e^{ln 5} ]^{- 2(log 10 base 5 - 1)}

x = 5^{- 2(log 10 base 5 - 1)}

x = 1/{5^2} ^{log 10 base 5 - 1}

x = 1/25 ^{log 10 base 5 - 1}

x = 1/25^{-1} × 1/25^{log 10 base 5}

x = 1/25^{-1} × (1/5^2)^{log 10 base 5}

x = 1/25^{-1} × 1/10^2

x = 25/100

∴ x = 1/4

In terms of mathematical expressions:

$\dfrac{1}{\log _x\left(10\right)}+2=\dfrac{2}{\log _5\left(10\right)}\\\\\mathrm{Rewrite\:the\:equation\:with\:}\log _x\left(10\right)=u\\\dfrac{1}{u}+2=\dfrac{2}{\log _5\left(10\right)}\\u=-\dfrac{\log _5\left(10\right)}{2\left(\log _5\left(10\right)-1\right)}\\\\\mathrm{Substitute\:back}\:u=\log _x\left(10\right)\\\log _x\left(10\right)=-\dfrac{\log _5\left(10\right)}{2\left(\log _5\left(10\right)-1\right)}$

$\dfrac{1}{\log _{10}\left(x\right)}=-\dfrac{\log _5\left(10\right)}{2\left(\log _5\left(10\right)-1\right)}\\1\cdot \:2\left(\log _5\left(10\right)-1\right)=-\log _{10}\left(x\right)\log _5\left(10\right)\\2\left(\log _5\left(10\right)-1\right)=-\dfrac{\ln \left(x\right)}{\ln \left(5\right)}\\\ln \left(x\right)=-2\ln \left(5\right)\left(\log _5\left(10\right)-1\right)\\x=e^{-2\ln \left(5\right)\left(\log _5\left(10\right)-1\right)}$

$\mathrm{Expand\:}e^{-2\ln \left(5\right)\left(\log _5\left(10\right)-1\right)}\\=\left(e^{\ln \left(5\right)}\right)^{-2\left(\log _5\left(10\right)-1\right)}\\=5^{-2\left(\log _5\left(10\right)-1\right)}\\=\dfrac{1}{5^{2\left(\log _5\left(10\right)-1\right)}}\\=\dfrac{1}{\left(5^2\right)^{\left(\log _5\left(10\right)-1\right)}}\\=\dfrac{1}{25^{\left(\log _5\left(10\right)-1\right)}}\\=\dfrac{1}{25^{\log _5\left(10\right)-1}}\\25^{\log _5\left(10\right)-1}=25^{\log _5\left(10\right)}\cdot \:25^{-1}$

$=10^2\cdot \:25^{-1}\\Now, \dfrac{1}{25^{\log _5\left(10\right)-1}} = \dfrac{25}{100} \\\therefore x=\dfrac{1}{4}$