Proving/manipulation question

**fterh** · September 10th 2010, 08:33 PM

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.

**Stroodle** · September 10th 2010, 08:56 PM

Are you sure you copied the question correctly? as:

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

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

So,

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

but this gives $-\frac{118}{125}=\frac{456}{25}$ , which is false.

**fterh** · September 10th 2010, 08:58 PM

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.

**TheCoffeeMachine** · September 10th 2010, 09:16 PM

$\displaystyle ({3^{x+2}})(5^{x-1}) = 15^{2x}$ $\displaystyle \Rightarrow (3^{x+2})(5^{x-1}) = 15^{x}\cdot 15^{x}$ $\displaystyle \Rightarrow \frac{3^x\cdot 3^2 \cdot 5^{x}\cdot 5^{-1}}{15^x} = 15^{x}$ $\displaystyle\Rightarrow \frac{9}{5}\left(\frac{3^{x}\cdot 5^{x}}{15^x}\right) = 15^{x}$ .

**fterh** · September 10th 2010, 09:17 PM

How do you know to do it this way rather than square rooting both sides?

**TheCoffeeMachine** · September 10th 2010, 09:23 PM

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 $x^{a+b} = x^ax^b$ , right?

**Prove It** · September 10th 2010, 09:27 PM

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.

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

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

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

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

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

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

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

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

**TheCoffeeMachine** · September 10th 2010, 09:48 PM

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.

**fterh** · September 10th 2010, 10:43 PM

Thanks a lot, gonna mark this as Solved. I guess the key is to practice and gain extensive exposure, huh.

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