1. ## Solve for X

Solve for x: log(2x+20)=-2

2. Originally Posted by VNVeteran
Solve for x: log(2x+20)=-2
hope this is right.

$\displaystyle log_{10} (2x + 20) = -2$

$\displaystyle 2x + 20 = \frac{1}{10^2}$

$\displaystyle 2x + 20 = \frac{1}{100}$

$\displaystyle 200x + 2000 = 1$

$\displaystyle x = -\frac{1999}{200}$

3. Hello
$\displaystyle \log(2x+20)=-2\Rightarrow 2x+20=\exp(-2)\Leftrightarrow x$$\displaystyle =\frac{\exp(-2)-20}{2}=\frac{1-20\exp(2)}{2\exp(2)}\approx -9,93233.$

4. Originally Posted by ukorov
hope this is right.

$\displaystyle log_{10} (2x + 20) = -2$

$\displaystyle 2x + 20 = \frac{1}{10^2}$

$\displaystyle 2x + 20 = \frac{1}{100}$

$\displaystyle 200x + 2000 = 1$

$\displaystyle x = -\frac{1999}{200}$
Same way I'd have done it.

5. Wrong answer Raoh , because log is a base 10 logarithm not base e , base e log is the " ln " so ukorov answer is the right answer

6. Originally Posted by mosta86
Wrong answer Raoh , because log is a base 10 logarithm not base e , base e log is the " ln " so ukorov answer is the right answer
well..he should have written $\displaystyle \log_{10}(2x+20)=-2$,i assumed $\displaystyle \log$ is the natural logarithm.

7. yes . but as it is by convention log is log base 10

8. Originally Posted by mosta86
yes . but as it is by convention log is log base 10
well..in where i live u should precise.

9. lol .. ok but check you calculator or any calculator , the log on it is base 10 , check all the references and books , log is base 10 ,, any other base you should specify . > Mostafa Ghalayini
B.E Computer and Communication engineering
M.S.E Computer and Communication engineering

10. didn't i say "I ASSUMED" ?
.END OF MY ASSUMPTION.

11. Actually, it is common practice in advanced math books (Calculus or higher) to use just "log" to mean natural logarithm since common logarithms are seldom if ever used. The practice in more basic books is use "log" to mean common logarithms and "ln" to mean natural logarithms. Here, the question was posted in "Pre-Calculus" so I would have assumed that common logarithms were meant, but it is certainly not true that "log" always means common logarithm.

In any case, it would have been better if the original post had specified which logarithm was intended and if the responders had said which convention they were using!