Question

2^x+1=3^x-1 find the x ​

Answer 1

$\bigstar$ Explanation $\bigstar$

$\leadsto$ Solution:-

$2^{x+1} = 3^{x-1}$

$\rm Taking\:log\:on\:both\:sides$

$\log(2^{x+1}) = \log(3^{x-1})\\\rm Use\:the\:property\:of\:logarithms\:that\: \log(a^x)=x log(a)\\(x+1)\log(2) = (x-1)log(3)\\\xlog(2) + log(2) = xlog(3) - log(3)$

Taking all the terms which contain only log function to one side and the others to the other side.

$\rm x\log(2) - x\log(3) = -\log(2) - \log(3)\\x[\log(2)-\log(3)] = -[\log(2)+\log(3)]$

Therefore,

$\rm x = \dfrac{-[log(2)+log(3)]}{[log(2)-log(3)]}$

We know that log 2 = 0.301 and log 3 = 0.477

Therefore,

$\rm x = \dfrac{-[0.301+0.477]}{[0.301-0.477]}$

$\rm x = \dfrac{-0.778}{-0.176} = \dfrac{0.778}{0.176} =4.42$

$\leadsto$ Important values of logarithms to remember:-

i) log 2 = 0.301

ii) log 3 = 0.477

iii) log 5 = 0.698

iv) log 7 = 0.845