Question 823929
{{{log(4, (3y)) - 2log(4, (x)) - 1 = 0}}}
First we will combine the logs. To do so we need to use a property of logarithms, {{{n*log(a, (p)) = log(a, (p^n))}}}, to "move" the 2 in front of the second log:
{{{log(4, (3y)) - log(4, (x^2)) - 1 = 0}}}
Now we can use another property of logs, {{{log(a, (p)) - log(a, (q)) = log(a, (p/q))}}}, to combine the logs:
{{{log(4, ((3y)/x^2)) - 1 = 0}}}<br>
Now that we are down to one log, we will isolate it. Adding 1 to each side:
{{{log(4, ((3y)/x^2)) = 1}}}
Next we rewrite the equation in exponential form. In general {{{log(a, (p)) = n}}} is equivalent to {{{p = a^n}}}. Using this pattern on our equation we get:
{{{(3y)/x^2 = 4^1}}}
which simplifies to:
{{{(3y)/x^2 = 4}}}<br>
Now that the logs are gone, we can solve for y. Multiplying both sides by {{{x^2}}}:
{{{3y = 4x^2}}}
Dividing by 3:
{{{y = (4x^2)/3}}}