We can use the change of base formula, , to change the two logs into logs that have the same base. Since both 4 and 256 and both powers of 4, I'm going to change the bases to 4's:
Since we chose a base for which 4 and 256 are powers, we can figure out the logs in the first denominator and second numerator:
Now we solve for . First we eliminate the fractions by multiplying both sides by the lowest common denominator:
which simplifies as follows:
The equation is in quadratic form for . If you have trouble seeing this then a temporary variable may help. Let
substituting z into the equation we get:
This is clearly a quadratic equation. To solve it we first want one side to be zero. Subtracting 3z from each side:
Now we factor:
From the Zero Product Property: or
Solving these we get: or
Of course these are solutions for z, our temporary variable. Now we need to substitute back for the temporary variable: or
And solve for x. If you can't see what values of x work in these equations, rewrite them in exponential form: or
which simplify to: or
Next we check these solutions. This is not optional when solving logarithmic equations like this. You must ensure that all arguments and bases are positive (and no base is a 1 either). Any "solution" that fails to make the arguments and bases valid must be rejected. Use the original equation to check:
Checking x = 256:
We can see that the arguments and bases are positive (and the bases are not 1's). So this solution passes the check.
Checking x = 1/4:
We can see that the arguments and bases are positive (and the bases are not 1's). So this solution also passes the check.
The problem asks for the product of these solutions. (This may be a clue to a simpler, faster solution which does not require finding the individual solutions for x.) So the answer to the problem is 256 * 1/4 = 64.
P.S. Once you have some experience with quadratic form equations you will not need a temporary variable. You will be able to see how to go from
to
to
to or
etc.
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