# Thread: 4^(2log(base8)7), equivalent expression, integers

1. ## 4^(2log(base8)7), equivalent expression, integers

Simplify 4^(2log(base8)7) into the form a^(b/c), all integers.

Could someone point me in the right direction?

2. Originally Posted by jsmith90210
Simplify 4^(2log(base8)7) into the form a^(b/c), all integers.

Could someone point me in the right direction?

First, convert 2Log8(7) > Log8(49)

Then, I believe you can use the following rule:
$
a^{\text{Logb}(c)}=c^{\text{Logb}(a)}
$

Therefore,
$
4^{\text{Log8}(49)}=49^{\text{Log8}(4)}
$

Log8(4) = 3/2

So, you have
49^(3/2)

---------------------------------
Josh
---------------------------------

3. Originally Posted by Invertible
First, convert 2Log8(7) > Log8(49)

Then, I believe you can use the following rule:
$
a^{\text{Logb}(c)}=c^{\text{Logb}(a)}
$

Therefore,
$
4^{\text{Log8}(49)}=49^{\text{Log8}(4)}
$

Log8(4) = 3/2 ...... Unfortunately no.

So, you have
49^(3/2)

---------------------------------
Josh
---------------------------------
......

4. Originally Posted by earboth
......
My mistake, I believe that:
Log8(4) = 2/3

$49^{(2/3)}$

5. what is the name of that rule?

6. Hello, jsmith90210!

A different approach . . .

Simplify $4^{2\log_87}$ into the form $a^{\frac{b}{c}}$, all integers.

We have: . $4^{\log_8(7^2)} \:=\:4^{\log_8(49)}$ .[1]

Let $\log_8(49) = p \quad\Rightarrow\quad 8^p \,=\,49\quad\Rightarrow\quad \left(2^3\right)^p \,=\,49 \quad\Rightarrow\quad 2^{3p} \,=\,49$

. . Then: . $3p \,=\,\log_2(49) \quad\Rightarrow\quad p \:=\:\tfrac{1}{3}\log_2(49)
$

. . Hence: . $\log_8(49) \:=\:\tfrac{1}{3}\log_2(49)$

Substitute into [1]: . $4^{\frac{1}{3}\log_2(49)} \;=\;\left(2^2\right)^{\frac{1}{3}\log_2(49)}$

. . . . . $=\;2^{\frac{2}{3}\log_2(49)} \;=\; 2^{\log_2(49^{\frac{2}{3}})} \;=\;\boxed{49^{\frac{2}{3}}}$

This answer is correct ... it satisfies the requirements of the problem.

But I bet that "they" simplified it further: . $49^{\frac{2}{3}} \:=\:\left(7^2\right)^{\frac{2}{3}} \:=\:7^{\frac{4}{3}}$