Question

Hi i just need a simple explanation for this question!

Answer 1

x =
`(5)/(3)`

Explanation:

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

These problems are getting you ready to work with exponential functions,

and ultimately with logarithms. The point here is that with variable exponents (notice the exponent is an x (on the right one) and not a number. Variable exponents can not be solved with regular algebra "rules" you need new ones.

The new ones will be Logs....

For now (until you learn logs) , you have to use some "tricks"

the "trick" in this problem is that you have to realize that 27 =
`3^(3)`...

with "common bases" this problem becomes trivial

`3^(5) =(3^(3) ) ^(x)`

so now the bases are the same and the equals sign suggests that

3x = 5

thus x =
`(5)/(3)`

Answer 2

5/3

Explanation:

3^5 = 27 ^x

Rewrite 27 and 3^3

3^5 = 3^3^x

We know that a^b^c = a^(b*c)

3^5 = 3^(3x)

The bases are the same so the exponents are the same

5 = 3x

Divide by 3

5/3 =x