OpenStudy (anonymous):

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

6 years ago
OpenStudy (anonymous):

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

6 years ago
hartnn (hartnn):

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

6 years ago
OpenStudy (anonymous):

yes :(

6 years ago
hartnn (hartnn):

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?

6 years ago
OpenStudy (anonymous):

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

6 years ago
OpenStudy (anonymous):

basically divide the differences?

6 years ago
OpenStudy (anonymous):

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

6 years ago
hartnn (hartnn):

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

6 years ago
OpenStudy (anonymous):

ohhh

6 years ago
OpenStudy (anonymous):

but there's no more x? 1+x/1-x = log2/log5 1 = log2/log5

6 years ago
OpenStudy (phi):

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

6 years ago
hartnn (hartnn):

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

6 years ago
OpenStudy (phi):

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)

6 years ago
OpenStudy (anonymous):

ohh.. i see. Thank you!

6 years ago
OpenStudy (phi):

so you got quite close to the answer...

6 years ago
OpenStudy (anonymous):

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

6 years ago
OpenStudy (phi):

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

6 years ago
OpenStudy (phi):

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

6 years ago
OpenStudy (phi):

using C= 0.4306

6 years ago
OpenStudy (anonymous):

ohh, I'm sorry :$thank you again! 6 years ago OpenStudy (phi): What were you doing to get the wrong answer? 6 years ago OpenStudy (anonymous): I guess I didn't even think about just putting 1+.4306 instead of inputting on my calculator: 1+log2/log5. What I forgot was inputting brackets so that messed it up :$

6 years ago
OpenStudy (phi):

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

6 years ago
OpenStudy (anonymous):

yup, thanks again!

6 years ago