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The equation $22\log(92x+40.66)=38.9$

steps so far

$\log(92x+40.66)=\frac{38.9}{22}$

to eliminate log, do I have to apply the opposite of log? Not sure what that is.

  • 0
    Yes, exponentiate both sides. Where did the minus sign in front of $\log$ come from?2012-12-05

3 Answers 3

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$22\log(92x+40.66)=38.9\Longrightarrow \log(92x+40.66)=\frac{38.9}{22}\Longrightarrow$

$92x+40.66=e^{\frac{38.9}{2}}\Longrightarrow \,\,\ldots$

If by $\,\log\,$ you mean logarithm in base $\,10\,$ just change $\,e\,$ for $\,10\,$

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In general, $\log_{b}u=v \iff b^v=u.$

Apply this fact. :-)

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You might have heard that $ b^{\log_b(x)} = x. $ Most people that I am aware of just write $\log$ when they mean $\log_{10}$. So that would mean that $ 10^{\log(x)} = x. $ Now you have the equation $ \log(92x+40.66)=\frac{38.9}{22} $ and to get rid of the $\log$ you can $ 10^{\log(92x+40.66)}=10^{\frac{38.9}{22}}. $ This is equivalent to $ 92x + 40.66 = 10^{\frac{38.9}{22}} $ and this is just a linear equation that you probably know how to solve (on both sides: subtract $40.66$ and then divide by $92$).