# Is this an unsolvable LOG equation?

#### Seventy7

Solve: log (x-10) = 3 + log (x-3)

Seems that no matter what is substituted for x, the log (x-10) is never able to equal the log (x-3) plus 3. Is this equation unsolvable, or are there theoretical/imaginary solutions?

#### romsek

Math Team
I'm assuming $\log(x) \sim \log_{10}(x)$

$\log(x-10)=3+\log(x-3)\\ \log(x-10)-\log(x-3) = 3 \\ \log\left(\dfrac{x-10}{x-3}\right) = 3\\ \left(\dfrac{x-10}{x-3}\right) = 10^3 = 1000\\ x-10 = 1000(x-3) = 1000x - 3000\\ 999x = 2990\\ x = \dfrac{2990}{999}$

• Seventy7 and idontknow

#### idontknow

For any equation you must be able to spot $$\displaystyle 3=\log10^3$$.
Then remove the logs both sides .

• Seventy7

#### greg1313

Forum Staff
Assuming base 10 logs,

$$\displaystyle \log(x-10)=3+\log(x-3)$$

$$\displaystyle \log(x-10)-\log(x-3)=3$$

$$\displaystyle \log\left(\frac{x-10}{x-3}\right)=3\quad(*)$$

$$\displaystyle \frac{x-10}{x-3}=1000$$

$$\displaystyle x-10=1000x-3000$$

$$\displaystyle 999x=2990$$

$$\displaystyle x=\frac{2990}{999}$$

If you use $(*)$ to verify, you avoid complex numbers.

Posted $\sim$ romsek.

• Seventy7

#### Seventy7

Is x=2990/999 (2.993) an imaginary answer? Plug in 2.993 into the left side of original equation (log x-10):

log (2.993-10)

= log (-7.007), but this seems unsolvable?

#### DarnItJimImAnEngineer

2990/999 is a real number, but the logs are both complex numbers. So, you can only find the answer using algebra, or using a numeric environment that handles complex numbers.

• romsek and Seventy7

#### romsek

Math Team
I should have gone back and verified that the solution didn't break any rules, which it does.
But OP needs to specify whether we are talking strictly reals or not.

Given it's high school math it almost certainly is.

• Seventy7

#### idontknow

$$\displaystyle re^{i\pi}=-r \;$$ ; $$\displaystyle \; \ln(-r)=\ln(r)+i\pi$$.

$$\displaystyle \log(-7.007 )=\dfrac{\ln(-7.007 )}{\ln10} =\dfrac{\ln(7.007)}{\ln10}+i\dfrac{\pi}{\ln10}$$.

• Seventy7

#### skipjack

Forum Staff
If "log" in the original post is the natural logarithm, there is a real solution.

• idontknow