$$7^{3x+1}=5^x$$
I am trying to solve this equation. I solved the equation and got what I believe to be the correct answer, but when I verify the answer it appears to be incorrect. Any idea why? Here is my work thus far:
$$7^{3x+1}=5^x$$
$$3x\log7 + 1\log7 = x\log5$$
$$3x\log7-x\log5 = -\log7$$
$$x = -\frac{\log7}{3\log7-\log5}$$
$$x = -0.460$$
When I make $x = -0.46$ in the original equation, the equation is not satisfied. Am I solving incorrectly or verifying incorrectly?
it seems you get it right
$$ 7^{3x+1}=5^{x}\\log(7^{3x+1})=log(5^{x})\\(3x+1)log7 =xlog 5\\x(3log7 -log 5)=-log7\\x=\frac{-log7}{3log7 -log 5}\\=\frac{-log7}{log7^{3} -log 5}$$
-0,46 is an approximation of the final result you found with logarithms. If you want to replace x in your initial equation, do it with the exact form, not with the approximate value.
By the way, both results are correct, this result should solve the equation.
-0,46 is an approximation, used for you to understand what's approximately the value of your number. It doesn't mathematically solve your equation.