Question

How do I solve this to the nearest thousandth: 5^(1+x) = 2^(1-x)?

Answer 1

So far:

log5^(1+x) = log2^(1-x)
(1+x)log=(1-x)log2
(1+x) = (1-x)log2/log5

Answer 2

yes,its correct so far...u have problem with next step??

Answer 3

yes :(

Answer 4

can u use calculator to find log2/log5 ? or it has to be kept in that way only??
also u know the rule of componendo-dividendo?heard of that?

Answer 5

log2/log5 = 0. 43068
and is it like this: log6-log7 = log6/log7?

Answer 6

basically divide the differences?

Answer 7

and the answer to this question is: – 0. 398 but I don't know how...

Answer 8

yes it is
now, \[\frac{1+x}{1-x}=0.4306\]
add 1 on both sides and try to simplify..

Answer 9

ohhh

Answer 10

but there's no more x?

1+x/1-x = log2/log5
1 = log2/log5

Answer 11

let c= 0.4306
then (1+x)/(1-x) = c
mult both sides by (1-x):
1+x = c-cx
solve for x

Answer 12

where did your denominator go??
\[\frac{1+x+1-x}{1-x}=\frac{2}{1-x}\]

Answer 13

when you got to

(1+x) = (1-x)log2/log5

or

(1+x) = (1-x)C (for ease of typing)
then
1+x= C-Cx
1+x+Cx= C
x+Cx= C-1

(1+C)x= (C-1)
x= (C-1)/(1+C)

Answer 14

ohh.. i see. Thank you!

Answer 15

so you got quite close to the answer...

Answer 16

wait.. it gives me the wrong answer.. when it's suppose to be – 0. 398 instead I get -.1386...

Answer 17

sounds like a loose nut behind the enter button. :)

Answer 18

C-1= -0.5694
C+1= 1.4306
(C-1)/(C+1)= -0.5694/1.4306= -0.3980

Answer 19

using C= 0.4306

Answer 20

ohh, I'm sorry :$ thank you again!

Answer 21

What were you doing to get the wrong answer?

Answer 22

I guess I didn't even think about just putting 1+.4306 instead of inputting on my calculator: 1+log2/log5. What I forgot was inputting brackets so that messed it up :$

Answer 23

ok, that makes sense. It is good to figure out how you went wrong, so you can fix the problem.

Answer 24

yup, thanks again!