Asked by jenny
how do i solve:
log4x^3 + log2x^1/2 = 8
please help me
log4x^3 + log2x^1/2 = 8
please help me
Answers
Answered by
drwls
You should use parentheses to indicate just what you are taking the log of. Is it 4, 4x or 4x^3?
I assume that you mean
log(4x^3) + log(2 x^1/2) = 8
which can be rewritten
log(2^2*x^3) + log 2 + (1/2) log x = 8
or
2log 2 + 3log x + log2 + (1/2) log x = 8
3log2 + (7/2)log x = 8
Use a calculator to solve that. What is the base of your log? It will matter.
I assume that you mean
log(4x^3) + log(2 x^1/2) = 8
which can be rewritten
log(2^2*x^3) + log 2 + (1/2) log x = 8
or
2log 2 + 3log x + log2 + (1/2) log x = 8
3log2 + (7/2)log x = 8
Use a calculator to solve that. What is the base of your log? It will matter.
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