Solution:
Given:
[tex]\log _22^{x+3}=\log _24(3)^x[/tex]
Since both sides have the same logarithm base, we equate the equation by canceling out the logarithm.
Hence,
[tex]\begin{gathered} \log _22^{x+3}=\log _24(3)^x \\ 2^{x+3}=4(3)^x \end{gathered}[/tex]
Applying the law of exponents,
[tex]x^{a+b}=x^a.x^b^{}[/tex]
Hence,
[tex]\begin{gathered} 2^{x+3}=4(3)^x \\ 2^x\times2^3=4(3^x) \\ 2^x\times8=4(3^x) \\ \frac{8}{4}=\frac{3^x}{2^x} \\ 2=\frac{3^x}{2^x} \\ 2\times2^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 \\ \frac{1+x}{x}=\frac{\log 3}{\log 2} \\ \frac{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=\frac{1}{0.585} \\ x=1.7094 \end{gathered}[/tex]
Therefore, the value of x is 1.7094
