Question 250052
{{{log(4, (x)) = log(8, (4x))}}}
The key to the solution is to recognize that 4 and 8 are both powers of 2. So we can rewrite each of these logarithms in terms of base 2 logarithms. You might be able to realize that since {{{4 = 2^2}}} and {{{8 = 2^3}}} and since logarithms are exponents, that {{{(1/2)log(2, (q)) = log(4, (q))}}} and {{{(1/3)log(2, (q)) = log(8, (q))}}}. If this is hard to understand then here is some Algebra to show it. We will use a temporary variable:
Let {{{z = log(4, (x))}}}
Rewriting this in exponential form we get:
{{{4^z = x}}}
Replace 4 with {{{2^2}}}:
{{{(2^2)^z = x}}}
Use the property of exponents, {{{(a^p)^q = a^(p*q)}}}:
{{{2^(2z) = x}}}
Find the base 2 logarithm of each side:
{{{log(2, (2^(2z))) = log(2, (x))}}}
Using a property of logarithms, {{{log(a, (p^q)) = q*log(a, (p))}}}, we can move the exponent out front:
{{{2z*log(2, (2)) = log(2, (x))}}}
Since {{{log(2, (2)) = 1}}}  by definition:
{{{2z = log(2, (x))}}}
Multiply both sides by 1/2:
{{{z = (1/2)log(2, (x))}}}
Replace our temporary variable with what it represents:
{{{log(4, (x)) = (1/2)log(2, (x))}}}
Similar logic shows that
{{{log(8, (4x)) = (1/3)log(2, (4x))}}}<br>
So we can write our equation using base 2 logarithms:
{{{(1/2)log(2, (x)) = (1/3)log(2, (4x))}}}
Now we can use the property of logarithms used earlier, in the other direction, to move the coefficients back into the arguments as exponents:
{{{log(2, (x^(1/2))) = log(2, ((4x)^(1/3)))}}}
Now that we have two base 2 logarithms that are equal, their arguments must be equal:
{{{x^(1/2) = (4x)^(1/3)}}}
To solve this we'll raise both sides to the 6th power. (You'll see why in a minute.)
{{{(x^(1/2))^6 = ((4x)^(1/3))^6}}}
which simplifies to
{{{x^3 = (4x)^2}}}
(See why we used 6 now?)
Now we simplify
{{{x^3 = 16x^2}}}
and solve. With a cubed term, the way to solve this is to get one side equal to zero and factor. Subtract {{{16x^2}}} from each side:
{{{x^3 - 16x^2 = 0}}}
Factor out the GCF (which is {{{x^2}}}):
{{{x^2(x -16) = 0}}}
Now we can use the Zero Product Property which says that this product can be zero only if one of the factors is zero. So:
{{{x^2 = 0}}} or {{{x-16 = 0}}}
Solving each of these we get
{{{x = 0}}} or {{{x = 16}}}<br>
With logarithmic equations we should always check our answers. We must make sure that the solutions do not make an argument to any logarithm zero or negative.
{{{log(4, (x)) = log(8, (4x))}}}
Checking x = 0:
{{{log(4, (0)) = log(8, (4(0)))}}}
As we can see, when x = 0 we get arguments that are zero so we have to reject x = 0 as a solution.
Checking x = 16:
{{{log(4, (16)) = log(8, (4(16)))}}}
which simplifies to
{{{log(4, (16)) = log(8, (64))}}}
which simplifies to
{{{2 = 2}}} Check!<br>
So the only solution is x = 16.