QuestionExpress in logarithmic form. a) $25=5^{2}$ b) $2^{3}=8$ c) $27^{1 / 3}=3$ d) $\frac{1}{36}=6^{-2}$

Studdy Solution

STEP 1

What is this asking? We need to rewrite these exponential equations in their equivalent logarithmic forms. Watch out! Remember the relationship between exponents and logarithms: $\log_b a = c$ means $b^c = a$ .
Don't mix up the base, exponent, and result!

STEP 2

1. Rewrite $25 = 5^2$ in logarithmic form.
2. Rewrite $2^3 = 8$ in logarithmic form.
3. Rewrite $27^{1/3} = 3$ in logarithmic form.
4. Rewrite $\frac{1}{36} = 6^{-2}$ in logarithmic form.

STEP 3

We're given $25 = 5^2$ .
Here, our base is $5$ , our exponent is $2$ , and our result is $25$ .

STEP 4

Remember, $\log_b a = c$ means $b^c = a$ .
Let's match things up! $b$ is like our $5$ , $c$ is like our $2$ , and $a$ is like our $25$ .

STEP 5

So, we can rewrite $25 = 5^2$ as $\log_5 25 = 2$ .
Awesome!

STEP 6

We have $2^3 = 8$ .
Our base is $2$ , the exponent is $3$ , and the result is $8$ .

STEP 7

Using the relationship $\log_b a = c$ which means $b^c = a$ , we see $b$ is $2$ , $c$ is $3$ , and $a$ is $8$ .

STEP 8

Therefore, $2^3 = 8$ becomes $\log_2 8 = 3$ .
Fantastic!

STEP 9

We're given $27^{1/3} = 3$ .
The base is $27$ , the exponent is $\frac{1}{3}$ , and the result is $3$ .
Fractional exponents?
No problem!

STEP 10

Remembering $\log_b a = c$ means $b^c = a$ , we identify $b$ as $27$ , $c$ as $\frac{1}{3}$ , and $a$ as $3$ .

STEP 11

So, $27^{1/3} = 3$ transforms into $\log_{27} 3 = \frac{1}{3}$ .
Keep it up!

STEP 12

We have $\frac{1}{36} = 6^{-2}$ .
Our base is $6$ , the exponent is $-2$ , and the result is $\frac{1}{36}$ .
Negative exponents?
We've got this!

STEP 13

Using our key relationship $\log_b a = c$ meaning $b^c = a$ , we see $b$ is $6$ , $c$ is $-2$ , and $a$ is $\frac{1}{36}$ .

STEP 14

Thus, $\frac{1}{36} = 6^{-2}$ becomes $\log_6 \frac{1}{36} = -2$ .
Excellent!

STEP 15

a) $\log_5 25 = 2$ b) $\log_2 8 = 3$ c) $\log_{27} 3 = \frac{1}{3}$ d) $\log_6 \frac{1}{36} = -2$

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