Answer:
4
Step-by-step explanation:
Here's my method of doing this:
We'll be using logarithms
Our question is:
[tex](\frac{1}{3})^{2}[/tex] × [tex](\frac{1}{3})^x[/tex] = [tex]\frac{1}{729}[/tex]
If we don't want to evaluate any powers at all we'll have to make each term have the same base (so that we can ignore the base — [tex]\frac{1}{3}[/tex])
we need to figure out what power of a [tex]\frac{1}{3}[/tex] the [tex]\frac{1}{729}[/tex] is.
what we can do is the following:
[tex](\frac{1}{3})^?[/tex] = [tex]\frac{1}{729}[/tex]
[tex]3^?[/tex] = 729
If we take the "log" of each, we can find out what power of 3 gives 729.[tex]log(3^?) = log\ 729[/tex]
We must remember this rule: [tex]log(a^n) = n \times log(a)[/tex]
[tex]? \times log 3 = log 729[/tex]
[tex]? = \frac{log\ 729}{log\ 3}[/tex]
? = 6 (with a calculator)
This means that to get 729, we need to do [tex]3^6[/tex]
Therefore [tex](\frac{1}{3})^6 = \frac{1}{729}[/tex]
We can now solve the equation!
[tex](\frac{1}{3})^{2}[/tex] × [tex](\frac{1}{3})^x[/tex] = [tex]\frac{1}{729}[/tex]
[tex](\frac{1}{3})^{2}[/tex] × [tex](\frac{1}{3})^x[/tex] = [tex](\frac{1}{3})^6[/tex]
Since each term of our equation has the same base of [tex]\frac{1}{3}[/tex] we can ignore it and just deal with the powers!
We must remember this rule: [tex]x^a \times x^b = x^{a + b}[/tex]
Using this rule:
[tex](\frac{1}{3})^{2}[/tex] × [tex](\frac{1}{3})^x[/tex] = [tex](\frac{1}{3})^6[/tex] becomes...
[tex]2 + x = 6[/tex]
[tex]x = 6 - 2 = 4[/tex]
