We want to solve each log equation.
Notice in each log equation, the unknown is the base.
To solve the log equations, we'll write
the corresponding exponential equation and then solve.
The first log equation is log base x of 27 equals three,
where x is the base, three is the exponent,
and 27 is a number.
Which means the corresponding exponential equation
is x raised to the third power or x cubed equals 27.
So to solve the equation x cubed equals 27,
we need to solve for x by undoing the cube.
One way to do this would be to take the cube root
of both sides of the equation.
So let's go ahead and do that.
We'll take the cube root of the left side
and the cube root of the right side.
The cube root of x cubed simplifies to one factor of x.
We have x equals the cube root of 27.
And 27 is equal to three times three times three
or three cubed, so we have x equals the cube root
of three times three times three or three cubed,
which simplifies perfectly to one factor of three,
giving us a solution of x equals three.
Next we have log base x of 256 equals four,
where x is the base, four is the exponent,
and 256 is a number.
The corresponding exponential equation
is x to the fourth power equals 256.
Now before we solve this, we need to remember
that the base of the logarithm has to be greater than zero,
and therefore we're only concerned
about the positive solutions to this equation.
And now to undo the fourth power,
we'll take the fourth root of both sides of the equation.
And we're taking the principle fourth root
because we're only concerned about the positive solution.
So the fourth root of x to the fourth
is equal to one factor of x.
We have x equals the fourth root of 256.
To simplify this,
let's look at the prime factorization of 256.
256 is equal to two times 128.
128 is equal to two times 64.
64 is equal to eight times eight.
Eight is equal to four times two.
And four is equal to two times two.
So we have one, two, three, four, five, six,
seven, eight factors of two.
So because the index is four, we're looking
for groups of four equal factors.
Here's two to the fourth
and here's another two to the fourth.
So we have x equals,
the fourth root of two to the fourth is equal
to one factor of two here and one factor of two here,
giving us x equals two times two or x equals four.
For the last log equation, x is the base,
negative four is the exponent,
and one over 625 is a number.
Which means the corresponding exponential equation
is x raised to the power of negative four
equals one over 625.
So again x to the negative fourth power
equals one over 625.
Remember, x to the power of negative four
is equal to one over x to the fourth.
Let's write this as one over x to the fourth
equals one over 625.
Well notice how if these two fractions
are equal to each other,
because the numerators are both one,
x to the fourth must equal 625.
Let's solve this equation to determine the solution
to the log equation, and again remember,
x has to be greater than zero.
So again we'll take the fourth root of both sides
of the equation to undo the fourth power
and solve for x.
The fourth root of x to the fourth is equal
to one factor of x,
giving us x equals the fourth root of 625.
Let's determine the prime factorization of 625.
Well 625 is equal to 25 times 25
and 25 is equal to five times five.
Notice 625 is equal to four factors of five.
Which means this will simplify perfectly
because we have the fourth root of five to the fourth
or four factors of five,
which simplifies to exactly one factor of five.
The solution is x equals five.
I hope you found this helpful.