# Math Help - indices problem

1. ## indices problem

Solve the equation $3^{x+3}-3^{x+2}=2$

2. I seem to be having trouble working out this one... can someone show me what I did wrong?

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

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

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

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

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

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

(By the way, x is a whole number)

3. Hi, DavdM, log is NOT linear function : log(a+b) is NOT equal to log(a)+log(b).

Here, we can write :

$3^{x+3}-3^{x+2}=2$
$3^{x+2}.(3-1)=2$
$3^{x+2}.2=2$
$3^{x+2}=1$
finally, $x=-2$

4. Originally Posted by thorin
$3^{x+2}.(3-1)=2$
I'm not getting where you got the (3-1) from. Please to explain?

5. $3^{x+3}-3^{x+2}=3.3^{x+2}-1.3^{x+2}=3^{x+2}.(3-1)$

6. Oh i get it.

7. Another way to do it: by the laws of exponents, $3^{x+ 3}= (3^x)(3^3)= 27(3^x)$ and $3^{x+ 2}= (3^x)(3^2)= 9(3^x)$. The equation becomes $3^{x+3}- 3^{x+ 2}= 27(3^x)- 9(3^x)= 18 (3^x)= 2$ so that $3^x= \frac{2}{18}= \frac{1}{9}$. Since $9= 3^2$, $\frac{1}{9}= 3^{-2}$ so we have, finally, $3^x= 3^{-2}$ and x= -2.