Question

the value of log 128base 4​

Answer 1

Answer:

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Answer 2

$1.\ \ \log_{b}a=\dfrac{\log a}{\log b}\\ \\ \\ 2.\ \ \log(a^b)=b\log a$

By using the identites given above:

$\begin{aligned}&\log_{4}128\\ \\ \Longrightarrow\ \ &\frac{\log 128}{\log 4}\\ \\ \Longrightarrow\ \ &\frac{\log \left(2^7\right)}{\log \left(2^2\right)}\\ \\ \Longrightarrow\ \ &\frac{7\log 2}{2\log 2}\\ \\ \Longrightarrow\ \ &\large \text{ \bold{\frac{7}{2}} }\end{aligned}$

Or we can do it without using logarithm!

$\begin{aligned}\textsf{Let}\ \ & \log_{4}128=x\\ \\ \Longrightarrow\ \ &4^x=128\\ \\ \Longrightarrow\ \ &(2^2)^x=2^7\\ \\ \Longrightarrow\ \ &2^{2x}=2^7\end{aligned}$

From this, we get,

$\begin{aligned}&2x=7\\ \\ \Longrightarrow\ \ &\large \text{ \bold{x=\frac{7}{2}} }\end{aligned}$

Hence 7/2 is the answer.