Question 608851
{{{5^log(5, (7))}}}
Once you really get what logarithms are, a problem like this is almost as easy as 1 + 1. So first let's look at the exponent: 
{{{log(5, (7))}}}
and try to understand what it represents.<br>
The idea behind logarithms is that it is possible to take any positive number (except 1) and, if you raise it to the right power, get any positive number as a result. For example, it is possible to raise 4 to some power and get 16. In fact you should know what the "right power" is: 2. It also possible to raise 4 to the "right power" and get 17! This exponent is not well-known but it exists. Similarly we can raise 12000 to some power and get 1/3, raise 51.3 to some power and get 1000, etc.<br>
In general, logarithms are exponents. {{{log(a, (p))}}} is how we express the "right exponent" for a to get p. For example:
{{{log(4, (16))}}} represents the exponent for 4 that results in 16.
{{{log(4, (17))}}} represents the exponent for 4 that results in 17.
{{{log(12000, (1/3))}}} represents the exponent for 12000 that results in 1/3.
{{{log(51.3, (1000))}}} represents the exponent for 51.3 that results in 1000.
etc.<br>
Some exponents are well-known:
{{{log(4, (16)) = 2}}} since 2 is the exponent for 4 that results in 16
{{{log(2, (64)) = 6}}} since 6 is the exponent for 2 that results in 64
{{{log(25.93, (1)) = 0}}} since <i>any number</i> raised to the 0 power is 1
{{{log(25, (5)) = 1/2}}} since 5 is the square root of 25 and an exponent of 1/2 means square root.
etc.<br>
Many logarithms are not well-known:
{{{log(4, (17))}}}
{{{log(12000, (1/3))}}}
{{{log(51.3, (1000))}}}
etc.
For these, if we need to see a number for the exponent for some reason, we have to reach for our calculators.<br>
Now let's look at your logarithm:
{{{log(5, (7))}}}
It represents the exponent for 5 that results in a 7. This is not a well-known... But look at where we see it! It's the exponent for a 5. So your full expression is "5 raised to the power for 5 which results in a 7". So:
{{{5^log(5, (7)) = 7}}}!!!<br>
In fact, the base number doesn't matter.
{{{a^log(a, (p)) = p}}}
no matter what "a" and "p" are (as longs as they are positive and a is not 1). It's like:<ol><li>Pick a positive number (not 1).</li><li>Now raise it to a power that we know will result in "p" (another positive number).</li><li>What answer do you get? "p", of course!</li></ol>