Question

Simplify the ratio

Answer 1

Can you give the ratio that you need to simplify?

Answer 2

\[\huge \frac{ 2^{\log_2^{1/4^{a}}}-3^{\log_{27}\left( a ^{2}-1 \right)^{3}-2a} }{ 7^{4\log_{49}^{a}}-a-1 }\]

Answer 3

@ParthKohli

Answer 4

@ganeshie8 @mathslover

Answer 5

I'm not able to understand the question, but you have to use \(\large a^{\log b}= b^{\log a}\) for most of it.

Answer 6

yes

Answer 7

Then use it...

Answer 8

Follow what Parth said.

Answer 9

In the numerator i got -a-1

Answer 10

I'm not even able to understand what it says. ;-;

What's the first term in the numerator? Is that the log of \(1/4 a\) or \(1/4^a\)?

Answer 11

The latter one

Answer 12

The second term would become \(\left(\left(a^2 - 1\right)^3\right)^{\log_{27}3} = a^2 - 1\).

Answer 13

And wait, is it \(\frac{1}{4}a\) or \(\frac{1}{4a}\)?

Answer 14

Oh, then the first term would simplify to \(\dfrac{1}{4^a}\)

Answer 15

Well. you need to get your algebra skills here...

Answer 16

The denominator would become \(a^2 - a - 1\)

Answer 17

How did you simiplify the denominator

Answer 18

\[\large 7^{4\log_{49} a}\]\[\large = 7^{\log_{49}(a^4)}\]\[\large = \left(a^4\right)^{\log_{49}7}\]\[= a^{2}\]

Answer 19

how did you take a^4 down

Answer 20

\[\large \boxed{a^{\log b} = b^{\log a}}\]

Answer 21

Okay