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$$3^{1 + 2\log_3(y-x)} = 48$$ With this problem I have difficulty getting rid of the exponent.

$2\log_5(2y - x - 12) = \log_5(y-x) + \log_5(y + x)$

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I'm confused about what you tried to do. Did you try to take the log base $5$ of both sides? If so, why base $5$? The natural base of logarithm to use is $3$, as that's the base of the exponential in the problem. –  Jonah Sinick Nov 1 '12 at 4:51
    
As it is pointed out below, you don't even need to do this, as $3^{2 \log_{3}(y-x)}=3^{\log_{3}(x-y)^2}=(x-y)^2$ –  Alex Nov 1 '12 at 5:34

2 Answers 2

Hint: For the first equation, use $3^{a+b}=3^a \cdot 3^b$, then the definition of $\log_3$ is that $3^{\log_3 x}=$ what?

For the second, raise $5$ to the power of each side.

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went and tried to use base 10. ( log(2y - x - 12) + log(2y - x - 12))/log(5) = ( log(y-x) + log(y+x) )/ log(5) am not sure what you mean by the note there not sure how to make it work –  chuwee Nov 1 '12 at 4:23
    
@chuwee: for the right side, the sum of logs is the log of the product. Then the same property defining $\log_5$ that I quoted for $\log_3$ works. for the left side, $2 \log x=\log (x^2)$ –  Ross Millikan Nov 1 '12 at 4:29

$3^{1+2\log_3(y-x)}=3\cdot 3^{2\log_3(y-x)}=3\cdot (3^{\log_3(y-x)})^2=3(y-x)^2$ using $a^{\log_ax}=x$

$3(y-x)^2=48\implies (y-x)^2=16,y=x\pm 4$

From the 2nd equation, $(2y-x-12)^2=(y-x)(y+x)$ using $\log_x (ab)=\log_x a+\log_x b$ where $x >0,\ne1$

If $y=x+4,x=y-4,\{2y-(y-4)-12\}^2=\{y+y-4\}4$

or $(y-8)^2=4(2y-4),y^2-24y+80=0,y=20$ or $4$ using the solution of quadratic equation.

If $y=20,x=20-4=16;y=4,x=4-4=0$

Similarly, for $y=x-4$

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