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Please answer this with full accurate and simple steps :

log of 8 to the base (1/4) = x,
then find x

The answer is supposed to be -3/2 but i dont know how to get it

2006-07-29 16:49:56 · 7 answers · asked by Worst 2 in Science & Mathematics Mathematics

7 answers

You need to find the power that you can raise 1/4 to to get 8.

It will have to be a negative power because that will cause the fraction to flip.

For example,
(1/4)^ (-1) = 4

Since 8 is (sqrt(4))^3, 8 = 4^(3/2) = (1/4)^(-3/2)

Therefore (-3/2) is the power you raise (1/4) to to get 8, so it is the solution to the equation.

2006-07-29 16:57:25 · answer #1 · answered by mathsmart 4 · 2 1

=> log (2*2*2) base 1/2*1/2
=> 3 log 2 / -2 log 2
=> -3/2 log 2 base 2 [ log a base a = 1 ]
=> -3/2

2006-07-29 19:22:24 · answer #2 · answered by sreenivas k 2 · 0 0

log of 8 to the base (1/4) = x
==>log8/log0.25 = x
==> -1.5 or -3/2 = x

2006-07-29 19:58:44 · answer #3 · answered by Prakash 4 · 0 0

log_(1/4)(8) = x
is equivalent to
(1/4)^x = 8

By properties of exponents we have
(4^(-1))^x = 2^3
4^(-x) = 2^3
(2^2)^(-x) = 2^3
2^(-2x) = 2^3
This implies
-2x = 3
Solving for x gives
x = 3/-2 or -3/2.

There is probably a quickly, simpler way to get there but I'm a little tired and don't see it right now.

2006-07-29 17:03:38 · answer #4 · answered by IPuttLikeSergio 4 · 0 0

First write out the equation, your base is 1/4, or 0.25. And the answer is 8, you need to solve for the explonent, lets call it X.

8 = 0.25^x

Here are the steps to solve.

ln(8) = ln(0.25^x) = x*ln(0.25)

ln(8) / ln(0.25) = x

and so x = -1.5 or - 3/2

2006-07-29 16:55:41 · answer #5 · answered by SnowXNinja 3 · 0 0

log(1/4)8

(log(8))/(log(1/4)) = (-3/2)

2006-07-30 05:14:28 · answer #6 · answered by Sherman81 6 · 0 0

I told u the rule we use in the other question u asked about the same topic....just please look at it
so

x =log.0.25(8)
x = ln8 / ln 0.25
x = 2.08 / -1.39
x = -1.5 = - 3/ 2

NOTICE that
ln 0.25 < 0 because
0.25 < 1

2006-07-29 18:00:01 · answer #7 · answered by M. Abuhelwa 5 · 0 0

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