We want to solve the given log equation.
Notice how, for this equation,
we cannot just isolate the logarithm
and then write the log equation as an exponential equation.
Because here, we have the square of log base three of y.
And here, we have log base three of y
raised to the power of negative two.
For the first step,
we will apply the power property of logarithms, shown here,
where log base b of x raised to the power of n
equals n times log base b of x.
And this property only applies to log base three of y
raised to the power of negative two.
Because here, y is being raised
to the power of negative two.
We cannot apply the power property of logarithms
to the square of log base three of y
because the exponent of two is attached
to the entire logarithm, not just the y.
So applying the power property of logarithms
on the right side of the equation
gives us the square of log base three of y
minus 15
equals negative two
times log base three of y.
And now from here,
because we have the square of log base three of y here,
we will perform a substitution
to make it easier to recognize we are going
to solve this by factoring.
So we'll let u equal log
base three of y,
which means, on the left side of the equation,
we would have u squared minus 15
equals, on the right side, negative two u.
Notice by performing this substitution,
we can see we have a quadratic equation,
which we can now solve by factoring.
The first step is to set one side
of the equation equal to zero.
Here, we'll add two u to both sides,
which gives us u squared plus two u
minus 15 equals zero.
And now we will factor.
If the left side does factor,
it will factor into two binomial factors.
The factors of u squared are u and u.
The factors of negative 15 that add to positive two
are positive five and negative three,
giving us a factor of u plus five
and a factor of u minus three.
Now that we have factored,
we'll perform the substitution again
and replace u with log base three of y.
In terms of y, we have the quantity log
base three of y plus five
times the quantity log base three of y minus three.
And again, this product is equal to zero.
And therefore, the first factor of log
base three of y plus five must equal zero,
or the second factor of log
base three of y minus three must equal zero.
And now we'll solve these two log equations
by isolating the logarithm and then writing the log equation
as an exponential equation.
So for the first equation,
we will subtract five on both sides,
which gives us log base three of y equals negative five.
Or for the second equation, we add three to both sides,
which gives us log base three of y equals three.
And now we'll write the log equations
as exponential equations using the definition
of a logarithm, shown here,
where the base is b, the exponent is x,
and the number is n.
So for log base three of y equals negative five,
the corresponding exponential equation is three
raised to the power of negative five equals y.
And for the second log equation,
the corresponding exponential equation is three
raised to the power of three equals y.
Simplifying, three to the power of negative five is equal
to one over three raised to the power of positive five,
and three to the fifth is equal to 243,
which gives us y equals one over 243.
And for the second equation, three cubed is equal to 27,
giving us y equals 27.
So the given log equation has two solutions.
I hope you found this helpful.