# Proving/manipulation question

• Sep 10th 2010, 08:33 PM
fterh
Proving/manipulation question
Given that (3^(x+2))*(5^(x-1)) = 15^(2x), show that 15^x = 9/5.

I managed to reach: 15^x= [3*sqrt(5)*sqrt(3^x)*sqrt(5^x)]/5, but can't reach 9/5.
• Sep 10th 2010, 08:56 PM
Stroodle
Are you sure you copied the question correctly? as:

$\displaystyle 15x=\frac{9}{5}$

$\displaystyle \therefore x=\frac{3}{25}$

So,

$\displaystyle \left (3\times\frac{3}{25}+2\right )\left (5\times \frac{3}{25}-1\right )$ must equal $\displaystyle 152\times\frac{3}{25}$

but this gives $\displaystyle -\frac{118}{125}=\frac{456}{25}$, which is false.
• Sep 10th 2010, 08:58 PM
fterh
Sorry, the statement was altered during the posting process. I copied the original statement (with superscript) but during posting it was altered to normal text.

The wrong statement has been corrected.
• Sep 10th 2010, 09:16 PM
TheCoffeeMachine
$\displaystyle \displaystyle ({3^{x+2}})(5^{x-1}) = 15^{2x}$ $\displaystyle \displaystyle \Rightarrow (3^{x+2})(5^{x-1}) = 15^{x}\cdot 15^{x}$ $\displaystyle \displaystyle \Rightarrow \frac{3^x\cdot 3^2 \cdot 5^{x}\cdot 5^{-1}}{15^x} = 15^{x}$ $\displaystyle \displaystyle\Rightarrow \frac{9}{5}\left(\frac{3^{x}\cdot 5^{x}}{15^x}\right) = 15^{x}$.
• Sep 10th 2010, 09:17 PM
fterh
How do you know to do it this way rather than square rooting both sides?
• Sep 10th 2010, 09:23 PM
TheCoffeeMachine
Quote:

Originally Posted by fterh
How do you know to do it this way rather than square rooting both sides?

Which step are you referring to? I didn't take any square roots. You understand that $\displaystyle x^{a+b} = x^ax^b$, right?
• Sep 10th 2010, 09:27 PM
Prove It
Quote:

Originally Posted by fterh
Given that (3^(x+2))*(5^(x-1)) = 15^(2x), show that 15^x = 9/5.

I managed to reach: 15^x= [3*sqrt(5)*sqrt(3^x)*sqrt(5^x)]/5, but can't reach 9/5.

$\displaystyle 3^{x + 2}\cdot 5^{x - 1} = 15^{2x}$

$\displaystyle \ln{\left(3^{x + 2}\cdot 5^{x - 1}\right)} = \ln{\left(15^{2x}\right)}$

$\displaystyle \ln{\left(3^{x + 2}\right)} + \ln{\left(5^{x - 1}\right)} = 2x\ln{(15)}$

$\displaystyle (x + 2)\ln{3} + (x - 1)\ln{5} = 2x(\ln{3} + \ln{5})$

$\displaystyle x\ln{3} + 2\ln{3} + x\ln{5} - \ln{5} = 2x\ln{3} + 2x\ln{5}$

$\displaystyle 2\ln{3} - \ln{5} = x\ln{3} + x\ln{5}$

$\displaystyle 2\ln{3} - \ln{5} = x(\ln{3} + \ln{5})$

$\displaystyle \frac{2\ln{3} - \ln{5}}{\ln{3} + \ln{5}} = x$.
• Sep 10th 2010, 09:48 PM
TheCoffeeMachine
Quote:

Originally Posted by fterh
How do you know to do it this way rather than square rooting both sides?

I think I misread your question earlier. Apologies. The answer is because taking the square root of both sides is not going to take you anywhere.
There is no method of identifying which technique to use what time; it's just insights that one gains from exposure and experience by doing lots of this.
• Sep 10th 2010, 10:43 PM
fterh
Thanks a lot, gonna mark this as Solved. I guess the key is to practice and gain extensive exposure, huh.