# Solve $33(1.207)^x = 47(1.547)^x$

How can we solve this equation? $$33(1.207)^x = 47(1.547)^x$$

I've done this step so far.$$33x\ln(1.207) = 47x\ln(1.547)$$

This is where I get stuck. Won't moving an $x$ to one side eliminate all the $x$'s?

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The step you did was wrong. From $ab^c=de^c$ you should get $\log a+c\log b=\log d+c\log e$. – Gerry Myerson Dec 4 '12 at 23:55

When you take the log of an each side of your equation, you need to remember that

"the log of a product is the sum of the logs":

$$\ln \left(a \cdot b^x \right) = \ln a + \ln (b^x) = \ln a + x \ln b$$

So what does that mean for your equation?

$$33(1.207)^x = 47(1.547)^x$$ $$\ln[33(1.207^x)] = \ln[47(1.547^x)]$$ $$\ln{(33)} + x \ln{(1.207)} = \ln{(47)} + x\ln{(1.547)}$$

Now solve for $x$

Gather constants to the right, multiples of $x$ to the right.

$$x \ln{(1.207)} - x\ln{(1.547)}= \ln{(47)} -\ln{(33)}$$

Factor out $x$:

$$x(\ln(1.207) - \ln{(1.547)}) = \ln{(47)} - \ln{(33)}$$

Now, divide both sides of the equation by $(\ln(1.207) - \ln(1.547))$, and you are left with:

$$x = \frac{\ln{(47)} -\ln{(33)}}{\ln(1.207) - \ln(1.547)}$$

Then simplify the expression using the property of logs: $$\ln(a) - \ln(b) = \ln\left(\frac ab\right)$$

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I like how you set up this explanation, but I still don't see how to get x by itself. I know you divide, but then you will have (xln(1.207))/(xln(1.547)), right? – Tyler Zika Dec 5 '12 at 0:09
Tyler: Does that help? – amWhy Dec 5 '12 at 0:17
my mind is blown.. yes that helps, thank you! – Tyler Zika Dec 5 '12 at 0:20

Recall that $$\log \left(a \cdot b^x \right) = \log a + x \log b$$ Hence, in your case, you will get that $$\log (33) + x \log(1.207) = \log (47) + x \log(1.547)$$

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Method 1: From $33(1.207)^x=47(1.547)^x$, we have $\ln 33+x\ln1.207=\ln 47+x\ln 1.547$, so that $x\ln 1.547-x\ln1.207=\ln 33-\ln 47$. Hence $x=\frac{\ln 33-\ln 47}{\ln 1.547-\ln 1.207}$.

Method 2: From $33(1.207)^x=47(1.547)^x$, we have $\frac{33}{47}=(\frac{1.547}{1.207})^x$, so that $x\ln\frac{1.547}{1.207}=\ln\frac{33}{47}$. Hence $x=\ln\frac{33}{47}/\ln \frac{1.547}{1.207}$.

Of course they can be seen to give the same answer.

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