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3 = log8 + 3logx. Solve for x.
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i dont no what it is
Me either
\(\large 3=log8+3logx \) \(\large 3=log8+log(x^3) \) \(\large 3=log(8\cdot x^3) \) now write that last line in exponential form... can you do that?
Would it be 8^x3=3?
no... that problem is the common log right? base 10?
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Yes it is
so where's the 10 in your equation?
It's not shown. I thought there would be one but that's all it had :/
\[\huge log_bM=x \Rightarrow b^x=M \]
your base is 10, and your exponent is the 3. the M is whatever is in the parenthesis...
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Okay
I think I got how to do it
so your equation in exponential form should be: \(\large 10^3=8x^3 \)
ok.... good luck to you...
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