Question 682019
{{{log(256, (x)) - log(x,(4)) = 3/4}}}
I've never done a problem like this. So there may be a better way than...<br>
We can use the change of base formula, {{{log(a, (p)) = log(b, (p))/log(b, (a))}}}, 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:
{{{log(4, (x))/log(4, (256)) - log(4,(4))/log(4, (x)) = 3/4}}}
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:
{{{log(4, (x))/4 - 1/log(4, (x)) = 3/4}}}<br>
Now we solve for {{{log(4, (x))}}}. First we eliminate the fractions by multiplying both sides by the lowest common denominator: {{{4log(4, (x))}}}
{{{4log(4, (x))(log(4, (x))/4 - 1/log(4, (x))) = 4log(4, (x))(3/4)}}}
which simplifies as follows:
{{{log(4, (x))*log(4, (x)) - 4*1 = log(4, (x))*3}}}
{{{(log(4, (x)))^2 - 4 = 3log(4, (x))}}}
The equation is in quadratic form for {{{log(4, (x))}}}. If you have trouble seeing this then a temporary variable may help. Let
{{{ z = log(4, (x))}}}
substituting z into the equation we get:
{{{z^2 - 4 = 3z}}}
This is clearly a quadratic equation. To solve it we first want one side to be zero. Subtracting 3z from each side:
{{{z^2 -3z - 4 = 0}}}
Now we factor:
{{{(z-4)(z+1) = 0}}}
From the Zero Product Property:
{{{z-4 = 0}}} or {{{z+1 = 0}}}
Solving these we get:
{{{z = 4}}} or {{{z = -1}}}<br>
Of course these are solutions for z, our temporary variable. Now we need to substitute back for the temporary variable:
{{{log(4, (x)) = 4}}} or {{{log(4, (x)) = -1}}}
And solve for x. If you can't see what values of x work in these equations, rewrite them in exponential form:
{{{4^4 = x}}} or {{{4^(-1) = x}}}
which simplify to:
{{{256 = x}}} or {{{1/4 = x}}}<br>
Next we check these solutions. This is <i>not optional</i> 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:
{{{log(256, (x)) - log(x,(4)) = 3/4}}}
Checking x = 256:
{{{log(256, (256)) - log(256,(4)) = 3/4}}}
We can see that the arguments and bases are positive (and the bases are not 1's). So this solution passes the check.<br>
Checking x = 1/4:
{{{log(256, (1/4)) - log(1/4,(4)) = 3/4}}}
We can see that the arguments and bases are positive (and the bases are not 1's). So this solution also passes the check.<br>
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.<br>
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
{{{(log(4, (x)))^2 - 4 = 3log(4, (x))}}}
to
{{{(log(4, (x)))^2 - 3log(4, (x)) -4 = 0}}}
to
{{{(log(4, (x)) - 4)(log(4, (x)) + 1) = 0}}}
to
{{{log(4, (x)) - 4 = 0}}} or {{{log(4, (x)) + 1 = 0}}}
etc.