find the value of x in the equation without evaluating the power

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]

Please help asap!! mark brainlyest.

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