Question

Question #ec244

Answer 1

`5^(3x) = 1/27`

Explanation:

We know that `color(red)(5^-x) = color(blue)3`

Remember the rule of exponents:

`(a^b)^c = a^(bc)`

Therefore:

`(color(red)(5^-x))^(-3) = 5^((-x)(-3)) = 5^(3x)`

So we need to raise `5^(-x)` to the "-3"rd power to get `5^(3x)`.

`(color(red)(5^-x))^(-3) = color(blue)3^(-3) = 1/color(blue)3^3 = 1/27`

Final Answer

Answer 2

`5^(3(-log(3)/log(5)))`
Or 0.37037....

Explanation:

In order to solve this, you can use a calculator or just use `log`.

However....if you don't know what `log` is, then allow me to give you a quick demonstration.
`log_a a^(b)=b`

This might seem a little confusing at first, but the a is the "base". B is supposed to be the exponent of a in order to get `a^b`.

An actual example: `log_5 25=2`.

As you could see, 5 is the `a` in this situation and `b` is the exponent therefore the "middle number" is 25 (because `5^2=25`).

Rule: If there is no base or a, then the a value is automatically assumed to be 10.

Now, another essential rule is that if a^b has an actual exponent on it (such as `log 100^2`), we can move that exponent to the left of log so it multiplies the result of log 100.

Proof: `log 100^2` or `log 10000`.
`log 10000=4` (because `10^4=10000`).

Now, let's multiply the result by two...

`2log 100`

`log 100=2`

Let's multiply the result by two...

`2*2=4`! Therefore it works!

As you can see here,

Firstly, `log` both sides in order to get:
`log5^-x=log3`.

Then, due to the rules of logarithms, we can move the power of the `-x` so it multiplies `log 5`.

`(-x)log5=log3`.

Divide both sides by `log 5`

`-x=log3/log5`.

Then multiply both sides by negative one to get:

`x=-log3/log5`.

Now that we know what x is, we can plug it into the equation to get...

`5^(-3(log3/log5)`

You can simplify the logs with a calculator in order to get 0.37037....

Forgive me for this long answer.