SOLUTION
We want to solve
[tex]7^{2x+4}=2^{x-5}[/tex]
Taking logarithm of both sides, we have
[tex]\begin{gathered} \log 7^{2x+4}=\log 2^{x-5} \\ (2x+4)\log 7=(x-5)\log 2 \\ \text{expanding we have } \\ (2x)\log 7+(4)\log 7=(x)\log 2-(5)\log 2 \end{gathered}[/tex]
Collecting like terms we have
[tex]\begin{gathered} (2x)\log 7-(x)\log 2=-(4)\log 7-(5)\log 2 \\ x(2\log 7-\log 2)=-4\log 7-5\log 2 \\ \text{dividing both sides by }(2\log 7-\log 2),\text{ we have } \\ x=\frac{-4\log 7-5\log 2}{2\log 7-\log 2} \end{gathered}[/tex]
Hence the solution set expressed in terms of logarithm is
[tex]x=\frac{-4\log7-5\log2}{2\log7-\log2}[/tex]
Using a calculator to obtain a decimal approximation, we have
[tex]\begin{gathered} x=\frac{-4\log7-5\log2}{2\log7-\log2} \\ x=\frac{-3.3804-1.5051}{1.6902-0.3010} \\ x=\frac{-4.8855}{1.3892} \\ x=-3.51677 \\ x=-3.52 \end{gathered}[/tex]
Hence the answer is -3.52 to 2 decimal places
