1. ## Logarithmic Equations

I've been trying to solve this equation for more than an hour already and I still get the wrong answer:

log5 (x) - log25 (x) + log125 (x) = 5

The answer is 5^6 or 15,625

but I keep getting 5^(15/2) for some reason.

2. There's some pretty nasty log bases there. Have you tried to change base?

i.e. $\log_5x = \frac{\ln x}{\ln{5}}$

3. I tried solving the equation with base Ln, but i get lost with the fractions and I get a very ugly number, so to speak

4. Maybe you can work backwards given

$\log_5 5^6 - \log_{25} 5^6 + \log_{125}5^6$

$6\log_5 5 - \log_{25} (5^2)^3 + \log_{125}(5^3)^2$

$6\log_5 5 - 3\log_{25} 5^2 + 2\log_{125}5^3$

$6\log_5 5 - 3\log_{25} 25 + 2\log_{125}125$

$6-3+2=5$

5. I did that just to make sure the answer was correct, but I still didn't get anywhere

6. $\log_5 x - \log_{25} x + \log_{125} x = 5$

$\frac{\ln x}{\ln 5} - \frac{\ln x}{\ln 25} + \frac{\ln x}{\ln 125} = 5$

$\ln x \cdot \bigg( \frac{1}{\ln 5} - \frac{1}{\ln 25} + \frac{1}{\ln 125} \bigg)= 5$

Can you carry on from there now?

7. Hello, >_<SHY_GUY>_<!

$\log_5(x) - \log_{25}(x) + \log_{125}(x) \:=\: 5$

Using the Base-change Formula, we have:

. . . $\displaystyle \frac{\ln x}{\ln 5} - \frac{\ln x}{\ln25} + \frac{\ln x}{\ln125} \;=\;5$

. . . . $\displaystyle \frac{\ln x}{\ln 5} - \frac{\ln x}{\ln 5^2} + \frac{\ln x}{\ln5^3} \;=\;5$

. . . $\displaystyle \frac{\ln x}{\ln 5} - \frac{\ln x}{2\ln 5} + \frac{\ln x}{3\ln 5} \;=\;5$

. . $\displaystyle \frac{6\ln x - 3\ln x + 2\ln x}{6\ln 5} \;=\;5$

. . . . . . . . . . . . . . $\displaystyle \frac{5\ln x}{6\ln 5} \;=\;5$

. . . . . . . . . . . . . . $5\ln x \;=\;30\ln 5$

. . . . . . . . . . . . . . . $\ln x \;=\;6\ln 5$

. . . . . . . . . . . . . . . $\ln x \;=\;\ln 5^6$

. . . . . . . . . . . . . . . . $x \;=\;5^6$