# indices problem

• Aug 14th 2010, 12:56 AM
mastermin346
indices problem
Solve the equation $\displaystyle 3^{x+3}-3^{x+2}=2$
• Aug 14th 2010, 01:30 AM
DavidM
I seem to be having trouble working out this one... can someone show me what I did wrong?

$\displaystyle 3^{x+3}-3^{x+2}=2$

$\displaystyle \log{3^{x+3}}-\log{3^{x+2}}=\log{2}$

$\displaystyle (x+3)\log{3}-(x+2)\log{3}=\log{2}$

$\displaystyle x\log{3} + 3\log{3} - x\log{3} - 2\log{3} = \log{2}$

$\displaystyle x\log{3}- x\log{3} = \log{2}- 3\log{3} + 2\log{3}$

The x's cancel eachother out... and I don't know where went wrong.

(By the way, x is a whole number)
• Aug 14th 2010, 01:33 AM
thorin
Hi, DavdM, log is NOT linear function : log(a+b) is NOT equal to log(a)+log(b).

Here, we can write :

$\displaystyle 3^{x+3}-3^{x+2}=2$
$\displaystyle 3^{x+2}.(3-1)=2$
$\displaystyle 3^{x+2}.2=2$
$\displaystyle 3^{x+2}=1$
finally, $\displaystyle x=-2$
• Aug 14th 2010, 01:44 AM
DavidM
Quote:

Originally Posted by thorin
$\displaystyle 3^{x+2}.(3-1)=2$

I'm not getting where you got the (3-1) from. Please to explain?
• Aug 14th 2010, 01:49 AM
thorin
$\displaystyle 3^{x+3}-3^{x+2}=3.3^{x+2}-1.3^{x+2}=3^{x+2}.(3-1)$
• Aug 14th 2010, 02:01 AM
DavidM
Oh i get it.
• Aug 14th 2010, 04:18 AM
HallsofIvy
Another way to do it: by the laws of exponents, $\displaystyle 3^{x+ 3}= (3^x)(3^3)= 27(3^x)$ and $\displaystyle 3^{x+ 2}= (3^x)(3^2)= 9(3^x)$. The equation becomes $\displaystyle 3^{x+3}- 3^{x+ 2}= 27(3^x)- 9(3^x)= 18 (3^x)= 2$ so that $\displaystyle 3^x= \frac{2}{18}= \frac{1}{9}$. Since $\displaystyle 9= 3^2$, $\displaystyle \frac{1}{9}= 3^{-2}$ so we have, finally, $\displaystyle 3^x= 3^{-2}$ and x= -2.