# Thread: logarithmic help

1. ## logarithmic help

Use the relationship x = a^y ? y = loga x to solve each of the following:
log5 25 = x

log3 x = 0

How would I go about doing those? I believe the 5 and 3 are the base.

2. $log_{5}25=x$

$5^x=25 (\because log_{a}b=c\iff a^c=b)$

$5^x=5^2$

$\therefore x=2$ (comparing exponents)

$log_{3}0=x$

$3^x=0$
Which is impossible, so, no solution.

3. Originally Posted by Referos
$log_{5}25=x$

$5^x=25 (\because log_{a}b=c\iff a^c=b)$

$5^x=5^2$

$\therefore x=2$ (comparing exponents)
Exactly right.

$log_{3}0=x$

$3^x=0$
Which is impossible, so, no solution.
No, the problem was $log_3 x= 0$ which is the same as $x= 3^0= 1$.

4. D'oh! Of course, you are correct; I misread the equation.