# Simple Logarithm

• Jan 25th 2011, 02:04 AM
vivek_master146
Simple Logarithm
I have a problem solving a very simple logarithm problem.

$\displaystyle \log_8 128 = ?$

I know that the answer is 7/3 but i don't know how the answer is derived. Can someone explain all the step .

Thanks.
• Jan 25th 2011, 02:16 AM
CaptainBlack
Quote:

Originally Posted by vivek_master146
I have a problem solving a very simple logarithm problem.

$\displaystyle \log_8 128 = ?$

I know that the answer is 7/3 but i don't know how the answer is derived. Can someone explain all the step .

Thanks.

$\displaystyle 128=2^7$

$\displaystyle 8=2^3$

and:

$\displaystyle \log_8(128)=\dfrac{\log_2(128)}{\log_2(8)}$
CB
• Jan 25th 2011, 02:46 AM
Prove It
Quote:

Originally Posted by vivek_master146
I have a problem solving a very simple logarithm problem.

$\displaystyle \log_8 128 = ?$

I know that the answer is 7/3 but i don't know how the answer is derived. Can someone explain all the step .

Thanks.

$\displaystyle \displaystyle \frac{7}{3}$ is correct, because

$\displaystyle \displaystyle \log_8{128} = \log_8{(8^{\frac{7}{3}})} = \frac{7}{3}\log_8{8} = \frac{7}{3}$.
• Jan 25th 2011, 02:49 AM
DrSteve
You can also change to exponential form first and get a common base:

$\displaystyle \log_8 128 = x$

$\displaystyle 8^x=128$

$\displaystyle (2^3)^x=2^7$

$\displaystyle 2^{3x}=2^7$

$\displaystyle 3x=7$

$\displaystyle x=\frac{7}{3}$