Question

Im supposed to solve this equation using logarithms but I don't know how.

Answer 1

Given:

$3(4^x)=5^(x+1)$

To find:

The solution of the given equation.

Solution:

We have,

$3(4^x)=5^(x+1)$

It can be written as

$3((2^2)^x)=5^(x+1)$

$3((2^{2x})=5^(x+1)$      $[\because (a^m)^n=a^{mn}]$

Taking log on both sides.

$\log [3((2^{2x})]=\log 5^(x+1)$

$\log 3+ \log ((2^{2x})=\log 5^(x+1)$        $[\because \log (ab)=\log a+\log b]$

$\log 3+ 2x\log 2=(x+1)\log 5$      $[\because \log x^n=n\log x]$

Putting values of logarithms, we get

$0.477+ 2x(0.301)=(x+1)0.699$      $[\because \log 2 =0.301, \log 3=0.477, \log 5=0.699]$

$0.477+ 0.602x=0.699x+0.699$

$0.602x-0.699x=0.699-0.477$

$-0.097x=0.222$

Divide both sides by -0.097.

$x=\dfrac{0.222}{-0.097}$

$x=-2.288656$

$x\approx -2.289$

Therefore, the value of x is -2.289.