Question

Solve for x: 3^x-3^-x/3^x+3^-x=1/4

Answer 1

okay

Answer 2

Kinda confusing, should use parenthesis ><

Answer 3

theres no paranthesis is the problem lol. ill try to rewrite it with it

Answer 4

3^x-3^-x=3^x-(1/3^x) by exponential properties

Answer 5

I'm guessing you have to use natural log (ln) to get down the exponential

Answer 6

what i just did is the denominator

Answer 7

3™-3º
-------
3™+3º = 1/4 solve for x

™=x º=-x
i had to get creative

Answer 8

Soo, (3^x) - (3^-x) = 1/4
Natural log all sides
x ln 3 + x ln 3 = ln (1/4)
Divide both sides ln 3
x + x = ln (1/4) / ln (3)
2x = ln (1/4) / ln (3)
Divide both side by 2
x = (ln (1/4) / ln (3)) / 2

Answer 9

the first part of the equation is a fraction.

Answer 10

OOH, hmm

Answer 11

(3^2x-1/3^x)/(3^2x+1/3^x)=(3^3x-3^x)/(3^3x+3^x)
=(3^2x-1)/(3^2x+1)

Answer 12

(3^2x-1)=(3^x-1)(3^x+1) so you have to replace the the denominator by this equation and som(3^x+1) will cancel with numerator

Answer 13

you will be left with this equation which is (3^x-1)/(3^x+1)=(1/4)

Answer 14

then you take the natural log of both sides

Answer 15

r u good

Answer 16

?

Answer 17

hello

Answer 18

robtobey can you find any solution
and is my derivation right ? Does my derivation make sense?

Answer 19

i think i solved it

Answer 20

and i am on the right track as well

Answer 21

(3^x-1)/(3^x+1)=(1/4) you multiply both sides (3^x+1) which results in
3^x-1=3^x/4+(1/4) adding +1 to both sides and substracting (3^x/4) to both sides will result 3/4(3^x)=1/2 and doing a bit of algebraic you will have 3^x=3/2 and now using natural log to both sides you will have x In(30)=In(2/3)

Answer 22

and x=In(2/3)/In(3)

Answer 23

there you go madam tell if there is no sense step

Answer 24

im so confused on your answer.

Answer 25

\[\frac{3^x-3^{-x}}{3^{-x}+3^x}=\frac{1}{4} \]Using Mathemtica, I got the following value for x:\[x=\frac{-\text{Log}[3]+\text{Log}[5]}{2 \text{Log}[3]}=0.232486760358964 \]\[\frac{3^x-3^{-x}}{3^{-x}+3^x} \to 0.25 \]using the 15 digit version of x above.

Note: The problem expression simplifies to the following according to Mathematica:\[\frac{8}{1+9^x}=3\]

Answer 26

Take the log base 10 of each side and apply the "PowerExpand" function to the result.\[\text{Log10}\left[9^x\right]==\text{Log10}\left[\frac{5}{3}\right]\text{ // }\text{PowerExpand} \]\[\frac{2 x \text{Log}[3]}{\text{Log}[2]+\text{Log}[5]}=\frac{-\text{Log}[3]+\text{Log}[5]}{\text{Log}[2]+\text{Log}[5]}\]Solve the preceding for that little x in the numerator of the LHS.