Given:
[tex]3(4^x)=5^(x+1)[/tex]
To find:
The solution of the given equation.
Solution:
We have,
[tex]3(4^x)=5^(x+1)[/tex]
It can be written as
[tex]3((2^2)^x)=5^(x+1)[/tex]
[tex]3((2^{2x})=5^(x+1)[/tex] [tex][\because (a^m)^n=a^{mn}][/tex]
Taking log on both sides.
[tex]\log [3((2^{2x})]=\log 5^(x+1)[/tex]
[tex]\log 3+ \log ((2^{2x})=\log 5^(x+1)[/tex] [tex][\because \log (ab)=\log a+\log b][/tex]
[tex]\log 3+ 2x\log 2=(x+1)\log 5[/tex] [tex][\because \log x^n=n\log x][/tex]
Putting values of logarithms, we get
[tex]0.477+ 2x(0.301)=(x+1)0.699[/tex] [tex][\because \log 2 =0.301, \log 3=0.477, \log 5=0.699][/tex]
[tex]0.477+ 0.602x=0.699x+0.699[/tex]
[tex]0.602x-0.699x=0.699-0.477[/tex]
[tex]-0.097x=0.222[/tex]
Divide both sides by -0.097.
[tex]x=\dfrac{0.222}{-0.097}[/tex]
[tex]x=-2.288656[/tex]
[tex]x\approx -2.289[/tex]
Therefore, the value of x is -2.289.