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Solve for $y$. When expressed in simplest form, what familiar kind of equation results? $$\log y = 10 + 0.5x$$

For this question, I would get rid of log first right? So, I would get

$$\begin{align*} 10^{\log y}&= 10^{10 +0.5x}\\ y &= 10^{10 + 0.5x}? \end{align*}$$ Yea... I don't know what I'm doing.. I don't think that's the answer.

Can someone help explain?

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  • $\begingroup$ All your work so far is good. You can simplify it a little further using the property $a^{b+c}=a^ba^c$ $\endgroup$ – eepperly16 Jun 18 '15 at 0:47
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$y = 10^{10 + 0.5x} = 10^{10} \times 10^{0.5x} = 10^{10} \times (10^{0.5})^x = 10.000.000.000 \times (\sqrt{10})^x$

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You're on the right track. If I've surmised the goal of the question correctly, these are the manipulations you're supposed to do: $$\begin{align*} y &= 10^{10 + 0.5x} \\ y &= 10^{10}\cdot 10^{0.5x}\\ y &= 10^{10}\cdot e^{0.5\ln(10)x}\\ y &= Ce^{ax} \end{align*}$$ where $C=10^{10}$ and $a=\frac{1}{2}\ln(10)$. The form $y=Ce^{ax}$ represents exponential growth/decay. (See here for example.)

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