=
Since
Let's replace the right side by =
The power of on the left must equal
the power of on the right, so
x = 3
That's the solution.
Edwin
Answer by bucky(2189) (Show Source): You can put this solution on YOUR website! You can do this problem without using logarithms if you wish. The process for doing it
involves:
.
First: use the power rule of exponents to raise to the x power. By this rule the
exponent "x" is applied to both the numerator and the denominator so the given equation becomes:
.
.
But 1 raised to the "x" power is always 1 no matter what the value of "x" is. So the equation
simplifies to:
.
.
Notice that the numerators are equal ... both of them are 1. So for this equation to be
true, the denominators must also be equal. Therefore you can say that:
.
.
Six squared is 36 and if you multiply that by 6 you get 216. So you can say that
6 cubed = 216 so "x" must be 3 to indicate that 6 is cubed.
.
If you have to do the problem by using logarithms, then start with the given equation:
.
.
and take the logarithm of both sides:
.
.
Now apply two rules of logarithms. The first rule is that in a logarithm of a quantity
that is raised to an exponent, the exponent can be brought out to become the multiplier
of the logarithm. Applying that rule to the left side of the equation, makes the equation
become:
.
.
The second rule of logarithms to apply is the one that says the log of a fraction is
equal to the log of the numerator minus the log of the denominator. Applying that rule
to both sides in this problem gives:
.
.
Now use the fact that log(1) = 0 and substitute 0 for log(1) on both sides to get:
.
.
and this simplifies to:
.
.
If you multiply both sides of this equation by -1 you get rid of the minus signs and the
equation becomes:
.
.
Divide both sides of this equation by to solve for x. This division
leads to:
.
.
Now use your calculator to find that for base 10 logs and . Substitute these two values to get:
.
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And when you do this division on a calculator you get .
.
Hope this helps you to understand a little more about both logarithms and also about handling
exponents.
.