# Math Help - natural logs proof eqn

1. ## natural logs proof eqn

can somebody help.I have no idea where to start with this equation. I presume its actually quite easy but some concept I have forgotton. Q
If x satisfies $3^x.4^{2x+1}=6^{x+2}$, show that
$x=\frac{ln9}{ln8}$,,,,

2. Originally Posted by oxrigby
can somebody help.I have no idea where to start with this equation. I presume its actually quite easy but some concept I have forgotton. Q
If x satisfies $3^x.4^{2x+1}=6^{x+2}$, show that
$x=\frac{ln9}{ln8}$,,,,
first, take the natural logarithm of both sides:

$\ln\left(3^x4^{2x+1}\right)=\ln\left(6^{x+2}\right )$

Now recall that $\ln(ab)=\ln(a)+\ln(b)$

Thus, $\ln\left(3^x4^{2x+1}\right)=\ln\left(6^{x+2}\right )\implies \ln\left(3^x\right)+\ln\left(4^{2x+1}\right)=\ln\l eft(6^{x+2}\right)$

Now recall that $\ln(a^b)=b\ln a$

So, $\ln\left(3^x\right)+\ln\left(4^{2x+1}\right)=\ln\l eft(6^{x+2}\right)\implies x\ln\left(3\right)+(2x+1)\ln\left(4\right)=(x+2)\l n\left(6\right)$

Do you think you can take it from here and try to solve for x?

--Chris

3. Originally Posted by oxrigby
can somebody help.I have no idea where to start with this equation. I presume its actually quite easy but some concept I have forgotton. Q
If x satisfies $3^x.4^{2x+1}=6^{x+2}$, show that
$x=\frac{ln9}{ln8}$,,,,
Take logs,
$\ln (3^x \cdot 4^{2x+1} ) = \ln 6^{x+2}$
Thus,
$\ln 3^x + \ln 4^{2x+1} = \ln 6^{x+2}$
Therefore,
$x\ln 3 + (2x+1) \ln 4 = (x+2)\ln 6$
We get,
$x\ln 3 + 2x\ln 4 + \ln 4 = x\ln 6 + 2\ln 6$
Thus,
$x(\ln 3 + 2\ln 4 - \ln 6) = 2\ln 6 - \ln 4$
This gives,
$x\cdot \ln \frac{3\cdot 2^4}{6} = \ln \frac{6^2}{4}$
Thus we get,
$x\ln 8 = \ln 9 \implies x = \frac{\ln 9}{\ln 8}$