Question

I am looking help for the question below? What is the value of ‘x’?

Answer 1

Given:

`\log _22^(x+3)=\log _24(3)^x`

Since both sides have the same logarithm base, we equate the equation by canceling out the logarithm.

Hence,

`\begin{gathered} \log _22^(x+3)=\log _24(3)^x \\ 2^(x+3)=4(3)^x \end{gathered}`

Applying the law of exponents,

`x^(a+b)=x^a.x^b^{}`

Hence,

`\begin{gathered} 2^(x+3)=4(3)^x \\ 2^x*2^3=4(3^x) \\ 2^x*8=4(3^x) \\ (8)/(4)=(3^x)/(2^x) \\ 2=(3^x)/(2^x) \\ 2*2^x=3^x \\ 2^(1+x)=3^x \\ \text{Taking the logarithm of both sides,} \\ \log 2^(1+x)=\log 3^x \\ \text{Applying the law of logarithm,} \\ \log a^x=x\log a \\ \text{Then,} \\ \log 2^(1+x)=\log 3^x \\ (1+x)\log 2=x\log 3 \\ (1+x)/(x)=(\log 3)/(\log 2) \\ (1+x)/(x)=1.5850 \\ 1+x=1.585x \\ \text{Collecting the like terms,} \\ 1=1.585x-x \\ 1=0.585x \\ \text{Dividing both sides by 0.585,} \\ x=(1)/(0.585) \\ x=1.7094 \end{gathered}`

Therefore, the value of x is 1.7094