Solve for x;
$\log_{12}x=\frac{1}{2}\log_{12}9+\frac{1}{3}\log_{12}27$
The only thing throwing me off is the one third and one half, which my book does not say how to fix.
Asked
Active
Viewed 89 times
0
-
Hint: $\log x^y=y\log x$. – vadim123 May 02 '13 at 03:14
-
You can use the algebraic properties of logarithms to simplify the right side. – Jonas Meyer May 02 '13 at 03:14
-
and $\log_c a+\log_c b=\log_c ab$ – Ehsan M. Kermani May 02 '13 at 03:16
2 Answers
2
$$\log_{12}9=\log_{12}3^2=2\log_{12}3$$
Similarly, $$\log_{12}27=\log_{12}3^3=3\log_{12}3$$
So, $$\log_{12}x=2\log_{12}3=\log_{12}3^2=\log_{12}9\implies x=9$$
lab bhattacharjee
- 274,582
0
$\log_{12}x=\log_{12}9^{0.5}+\log_{12}27^{0.3333....}\implies\log_{12}3+\log_{12}3\implies \log_{12}9= \log_{12}x$. Therefore, $x=9$. Sorry for my horrible typing, I don't know how to use that program to make the writing nicer.
iostream007
- 4,529
Ovi
- 23,737
-
http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – vadim123 May 02 '13 at 03:24