Question

show that log base 8 1000= log base 2 10 algebraically.

Answer 1

Please help me out, I'm stuck. thanks:)

Answer 2

Suppose that \(\large\log_8 1000 = y\). Rewriting the equation in exponential form gives you \(\large8^y = 1000\implies 2^{3y} = 10^3\). Now, can you reverse this process to rewrite this exponential equation as a logarithmic equation with base 2?

Answer 3

log

Answer 4

Recall that \(\large \log_a (b^c) = c\log_a b\).

Answer 5

From \(\large 3\log_2 (10) = 3x\), you wanted to divide by 3 to get \(\large \log_2(10) = x\). Now, I noticed you changed the variable I used originally from \(y\) to \(x\); if we had left everything in terms of \(y\), we would still get the same result -- \(\large \log_2 (10) = y\), but now recall that I originally let \(\large \log_8 (1000) = y\). Hence, it should be clear that \(\large\log_8(1000) = y = \log_2(10)\), which is what you wanted to show.

I hope this makes sense!

Answer 6

@ChristopherToni Thank you :)