# Solve the following exponential equation

$$7^{3x+1}=5^x$$

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?

• If anyone wants to edit the equations for me, please do so.. I'm not sure how. – McB Nov 18 '14 at 16:04
• Looks right. Double check your verification step! – Simon S Nov 18 '14 at 16:07
• Okay, I double-checked the verification and got 0.476951 = 0.477378.. Is this close enough, since my answer was only to two decimal places? – McB Nov 18 '14 at 16:08
• Right. Lose decimal places and you shouldn't expect the equality to be exact anymore. – Simon S Nov 18 '14 at 16:15
• Makes sense. Thanks. – McB Nov 18 '14 at 16:16

it seems you get it right $$7^{3x+1}=5^{x}\\log(7^{3x+1})=log(5^{x})\\(3x+1)log7 =xlog 5\\x(3log7 -log 5)=-log7\\x=\frac{-log7}{3log7 -log 5}\\=\frac{-log7}{log7^{3} -log 5}$$