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