# Solve the equation(using logarithms)

I'm having trouble with this math problem:

Solve the equation(using logarithms)

$7^{2x+1} = 5^x$

Thanks!

• Take the log of both sides. What do you get? – Simon S Dec 2 '14 at 15:49
• The key here is that $\log(a^b)=b\cdot\log(a)$. – KSmarts Dec 2 '14 at 15:51

Note that:

• $7^{2x+1}=7\cdot(7^2)^x$

Therefore:

• $7^{2x+1}=5^x\implies$

• $7\cdot(7^2)^x=5^x\implies$

• $7\cdot49^x=5^x\implies$

• $7=\left(\frac{5}{49}\right)^x\implies$

• $x=\log_{\frac{5}{49}}7\approx-0.8525785$

To solve this, you should know the following formula: $$\log a^b = b\log a$$ Using the above identity: $$7^{(2x+1)} = 5^x$$ $$(2x+1)\log 7 = x\log 5$$ $$\frac{2x+1}{x} = \frac{\log 5}{\log 7}$$ $$2 + \frac{1}{x} = \frac{\log 5}{\log 7}$$ $$x = \frac{2\log 7}{\log 5 - 2\log 7}$$

You can get the values of $\log 5$ and $\log 7$ from logarithmic tables. Substitute them in the above equation and you'll get the final answer.

$\log(7^{2x+1})= \log(5^{x}) \Rightarrow (2x+1)/x = c$, where $c = \log_{7}5$. Thus, $x = \frac{1}{c-2}$.