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=(0.222)/(-0.097)`

`x=-2.288656`

`x\approx -2.289`

Therefore, the value of x is -2.289.