# Math Help - Solve for x...

1. ## Solve for x...

Hi,

I'm not getting this. I'm wrong - as usual. May I have some help please?

If $3(27)^x = 9^{2x\,+\,1}$, what is the value of x?

$81^x = 9^{2x\,+\,1}$

$x \log 81 = (2x + 1) \log 9$

$x \log 81 = 2x \log 9 + \log 9$

$x \log 81 - 2x \log 9 = \log 9$

$x(\log 81 - 2 \log 9) = \log 9$

$x = \frac{\log 9}{\log 81 - 2 \log 9}$

x = Undefined

2. I can't read the power on the RHS, but remember that $\displaystyle 27 = 3^3$ and $\displaystyle 9=3^2$.

Convert everything to powers of $\displaystyle 3$ and then you can equate the powers.

3. I don't think are any real answers for x.

Edit: Sorry, I edited it a minute after because I didn't think anyone would reply so soon.
The only solution I can find was -1. But by calculations I cannot seem to find any other answers.

4. $\displaystyle 3\cdot 27^x = 9^{2x+1}$

$\displaystyle 3(3^3)^x = (3^2)^{2x+1}$

$\displaystyle 3\cdot 3^{3x} = 3^{2(2x+1)}$

$\displaystyle 3^{3x+1} = 3^{4x+2}$

$\displaystyle 3x+1=4x+2$

$\displaystyle 1 = x+2$

$\displaystyle x= -1$.

5. Originally Posted by Prove It
I can't read the power on the RHS,...
Sorry, I fixed it.

Originally Posted by Prove It
but remember that $\displaystyle 27 = 3^3$ and $\displaystyle 9=3^2$.

Convert everything to powers of $\displaystyle 3$ and then you can equate the powers.
$3(27)^x = 9^{2x\,+\,1}$

$3(3^3)^x = (3^x)^{2x\,+\,1}$

$3(3^{3x}) = (3)^{4x\,+\,2}$

$3 (3x \log 3) = 4x + 2(\log 3)$

$9x \log 9 = 4x \log 3 + 2\log 3$

$9x \log 9 - 4x \log 3 = 2 \log 3$

$x(9 \log 9 - 4 \log 3) = 2 \log 3$

$x = \frac{2 \log 3}{9 \log 9 - 4 \log 3}$

$x \approx -1.797$

6. Originally Posted by Educated
The only solution I can find was -1. But by calculations I cannot seem to find any other answers.

Thanks.

7. Originally Posted by Hellbent
Sorry, I fixed it.

$3(27)^x = 9^{2x\,+\,1}$

$3(3^3)^x = (3^x)^{2x\,+\,1}$

$3(3^{3x}) = (3)^{4x\,+\,2}$

$3 (3x \log 3) = 4x + 2(\log 3)$

$9x \log 9 = 4x \log 3 + 2\log 3$

$9x \log 9 - 4x \log 3 = 2 \log 3$

$x(9 \log 9 - 4 \log 3) = 2 \log 3$

$x = \frac{2 \log 3}{9 \log 9 - 4 \log 3}$

$x \approx -1.797$
If you want to use logarithms, then at this step

$\displaystyle 3(3^{3x}) = 3^{4x+2}$

you need to take the logarithm of both sides

$\displaystyle \log{[3(3^{3x})]} = \log{(3^{4x+2})}$

$\displaystyle \log{3} + \log{(3^{3x})} = \log{(3^{4x+2})}$

$\displaystyle \log{3} + 3x\log{3} = (4x+2)\log{3}$

$\displaystyle (3x+1)\log{3} = (4x+2)\log{3}$

$\displaystyle 3x+1=4x+2$

Go from there...

8. Originally Posted by Prove It
$\displaystyle 3x+1=4x+2$

Go from there...
$3x - 4x = 2 - 1$

$-x = 1$

$x = -1$