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.
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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
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 $\displaystyle \displaystyle \log_8{128} = \log_8{(8^{\frac{7}{3}})} = \frac{7}{3}\log_8{8} = \frac{7}{3}$.
You can also change to exponential form first and get a common base: $\displaystyle \log_8 128 = x$ $\displaystyle 8^x=128$ $\displaystyle (2^3)^x=2^7$ $\displaystyle 2^{3x}=2^7$ $\displaystyle 3x=7$ $\displaystyle x=\frac{7}{3}$