0
$\begingroup$

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

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

$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – Jonah Sinick Nov 1 '12 at 4:51
  • $\begingroup$ As it is pointed out below, you don't even need to do this, as $3^{2 \log_{3}(y-x)}=3^{\log_{3}(x-y)^2}=(x-y)^2$ $\endgroup$ – Alex Nov 1 '12 at 5:34
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 $\endgroup$ – chuwee Nov 1 '12 at 4:23
  • $\begingroup$ @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, $2 \log x=\log (x^2)$ $\endgroup$ – Ross Millikan Nov 1 '12 at 4:29
0
$\begingroup$

$3^{1+2\log_3(y-x)}=3\cdot 3^{2\log_3(y-x)}=3\cdot (3^{\log_3(y-x)})^2=3(y-x)^2$ 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 $y=x-4$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.