# Solving an exponential equation with logoarithm laws

$$2(5)^x = 3^{x+1}$$

I am trying to solve for $x$ in the above equation. Is there a way to make the bases the same to solve? Can I simplify the left side to $10^x$? I'm really not sure where to start to be honest. Here's what I've tried:

$$x\log 10 = (x+1)\log3$$

$$x = \frac{\log10}{\log3}$$

$$x = 0.477121$$

I'm not confident with the answer.. Is it correct? Any ideas?

• is it $2\cdot5^x=3^{x+1}$ – Dr. Sonnhard Graubner Nov 18 '14 at 18:57
• 5 is in parentheses, but yes. – McB Nov 18 '14 at 18:58
• $$\log [2 (5^x)] = \log(2) + x \log(5)$$ Then solve for $x$. – Joel Nov 18 '14 at 18:58
• $2\cdot5^x\color{red}\ne10^x$! – rae306 Nov 18 '14 at 19:07

take the logaithm of both sides then you will get $\ln(2)+x\ln(5)=(x+1)\ln(3)$ this must be solved for $x$ we have $\ln(2)+x\ln(5)=x\ln(3)+\ln(3)$ thus we get $x(\ln(5)-\ln(3))=\ln(3)-\ln(2)$ therefore $x=\frac{\ln(3)-\ln(2)}{\ln(5)-\ln(3)}$
$$2\cdot5^x=3^{x+1}\\2\cdot5^x=3\cdot3^x\\(\frac{5}{3})^x=\frac{3}{2}\\x\log(\frac{5}{3})=\log(\frac{3}{2})\\x(\log(5)-\log(3))=\log(3)-\log(2)\\x=\frac{\log(3)-\log(2)}{\log(5)-\log(3)}$$