Question

Logarithm Question.

Solve for x :

\(\log_{\cfrac{3}{4}} \log_8 (x^2+7) + \log_{\cfrac{1}{2}} \log_{\cfrac{1}{4}} (x^2+7)^{-1} = - 2\)

Answer 1

THis is what I have done yet :

\(\boxed{\log_\cfrac{3}{4} \left[ \cfrac{1}{3}\log_2 (x^2 +7) \right] - \log_2 \left[ \cfrac{1}{2} \log_2 (x^2 + 7) \right] \\}\)

Answer 2

The rule I used yet :

\(\large \boxed {\log_{a^q} (x) = \cfrac{1}{q} \log_a x}\)

Answer 3

@myininaya @mathmale

Answer 4

Changing the base will work?

Answer 5

Well, the second term can be simplified, I think :


\(\log_\cfrac{1}{2} \cfrac{1}{2} + \log_\cfrac{1}{2}\left( \log_2 (x^2 + 7) \right) \)

Answer 6

Let \(\log_2(x^2 + 7) = t\)

Answer 7

Okay,

1 + \(\log_\cfrac{1}{2} t \)

Answer 8

No!

Answer 9

So, I get :

\(\log_\cfrac{3}{4} \left(\cfrac{1}{3} t \right) + 1 + \log_\cfrac{1}{2} t = -2 \)

Answer 10

The second term is log base 2 right?

Answer 11

NVM.\[\boxed{\log_\cfrac{3}{4} \left[ \cfrac{1}{3}\log_2 (x^2 +7) \right] - \log_2 \left[ \cfrac{1}{2} \log_2 (x^2 + 7) \right] \\}\]Simplify the second term to \(-(1 + \log_2 t) \)

Answer 12

And the first term to\[\log_{3/4} 1/3 + \log_{3/4} t\]

Answer 13

Yeah, whoops.

Answer 14

The second term is \(-\left(-1 + \log_2 t\right)\)

Answer 15

Sorry, my net got down... Well yeah, I happened to make some confusion here.

Answer 16

\[\log_{3/4} (1/3) + \log_{3/4}(t) + 1 - \log_2 t = 2\]Yeah, we're almost done.

Answer 17

Write \(1/3\) as \(3^{-1}\).

Answer 18

@mathslover Can you continue?

Answer 19

Too many things to follow. Wait, I'll try to do this on paper and will return to you.

Answer 20

\[\log_{3/4}(t/3) - \log_2 (t/2)=-2\]\[\log_{3/4}t - \log_{3/4}3 - \log_2 t + 1 = -2\]\[\log_{3/4} t - \log_2 t = -3 + \log_{3/4} 3\]

Answer 21

I would try to get a common base on the LHS, but... lol

Answer 22

Yeah make it into a single fraction, and it's simple from there, kind of...you'll get a complex number, I'm sure.

Answer 23

\(-\log_\cfrac{3}{4} 3 + \log_2 t . \log_\cfrac{3}{4} 2 - \log_2 t = -3 \)

Answer 24

Use `\cfrac` only for continued fractions...

Answer 25

(guys, my net is not working properly, so, sorry for late responses)

And fromt he above equation :

\(\color{blue}{\text{Originally Posted by}}\) @mathslover
\(-\log_\cfrac{3}{4} 3 + \log_2 t . \log_\cfrac{3}{4} 2 - \log_2 t = -3 \)
\(\color{blue}{\text{End of Quote}}\)

We can continue like :

\(-\log_\cfrac{3}{4} 3 + \log_2 t ( \log_\cfrac{3}{4} 2 - 1 ) = -3\)

or

\(\log_2 t (\log_\cfrac{3}{4} 2 - 1) = \log_\cfrac{3}{4} 3 -3\)

Answer 26

Ah, nice :)

Answer 27

\(\log_2 t \left( \log_\cfrac{3}{4} \cfrac{8}{3}\right) = \log_\cfrac{3}{4} \left(\cfrac{3}{4}\right)^{-3} + \log_\cfrac{3}{4} 3 \)

Answer 28

Seems like it is almost done..

Answer 29

\(\log_2 t \times \log _ {3/4} \cfrac{8}{3} = \log_{3/4} \cfrac{3^{-2}}{4^{-3}} \\

\log_2 t \times \log_{3/4} \cfrac{8}{3} = \log_{3/4} \left(\cfrac{3^{-1}}{2^{-3}}\right)^2 \\

\log_2 t \times \log_{3/4} \cfrac{8}{3} =2 \log_{3/4} \cfrac{8}{3} \)

Answer 30

\(\log_2 t \times \cancel{\log_{3/4} \cfrac{8}{3}} = 2\cancel{\log_{3/4}{ \cfrac{8}{3}}} \)

\(\log_2 t = 2\)

Answer 31

Phew, at least I got close to solving it :)

Answer 32

Wooh...

t = 4

putting t = \(\log_2 {x^2 + 7}\)

\(\log_2 {x^2 + 7} = 4\)

\(x^2 + 7 = 16\)

\(x^2 = 9\)

Thus

\(x = \pm 3\)

Answer 33

Yeah @ParthKohli ... Thanks for your help.