Question

How to express exponential equation, 3^c = 27 , as a logarithmic equation

Answer 1

`log_(3)`27 = c

using the law of logs :
`log_(b)` x = n ⇔ x =
`b^(n)`

given
`3^(c)` = 27 then

`log_(3)` 27 = c

Answer 2

If you consider the logarithm base 3 of both sides, you have

`\log_3(3^c) = \log_3(27)`

You can use a rule of logarithms that allow you to turn exponents into multiplicative factors:

`\log_a(b^c) = c\log_a(b)`

So the equation becomes

`c\log_3(3)=\log_3(27)`

Now, by definition, you have

`\log_a(b)=c \iff a^c=b`

So, you have

`\log_3(3)=x \iff 3^x=3 \iff x = 1,\quad \log_3(27)=y \iff 3^y=27\iff y=3`

So, the equation becomes

`1\cdot c = 3 \iff c=3`