0
$\begingroup$

I'm having trouble with this math problem:

Solve the equation(using logarithms)

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

Thanks!

$\endgroup$
  • 2
    $\begingroup$ Take the log of both sides. What do you get? $\endgroup$ – Simon S Dec 2 '14 at 15:49
  • $\begingroup$ The key here is that $\log(a^b)=b\cdot\log(a)$. $\endgroup$ – KSmarts Dec 2 '14 at 15:51
0
$\begingroup$

Note that:

  • $7^{2x+1}=7\cdot(7^2)^x$

Therefore:

  • $7^{2x+1}=5^x\implies$

  • $7\cdot(7^2)^x=5^x\implies$

  • $7\cdot49^x=5^x\implies$

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

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

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.