need help with solving equation: (2^x - 2^-x) / 3 = 4

[jbizhop]

Hi! I am new to the site, so I'll do my best at explaining my problem.

. . .(2^x - 2^-x) / 3 = 4

In english, two raised to the x, minus two raised to the negative x, all divided by three, equals four.

First I multiply both sides by 3 to get rid of the 3 in the denominator and get:

. . .2^x - 2^-x = 12

This is equivlent to:

. . .2^x - (1/2^x) = 12

so I multiplied both sides by 2^x and get:

. . .2^2x - 1 = 12*2^x

Then I set the equation to zero getting:

. . .2^2x - 1 - (12*2^x) = 0

This is where I get stuck. I know at some point you need to use logs and I think before that you might substitute a letter for a piece of the equation. I am not sure. Please help. Thank you!

 

[JakeD]

Re: need help with this problem

hi I am new to the site so i'll do my best at explaining my problem.
(2^x - 2^-x)/3=4 In english, two raised to the x minus two raised to the negative x, all divided by three equals four. First i multiply both sides by 3 to get rid of the 3 in the denominator and get 2^x-2^-x=12. this is equivlent to 2^x - (1/2^x)=12, so i multiplied both sides by 2^x and get 2^2x-1=(12*2^x). Then i set the equation to zero getting 2^2x-1-(12*2^x)=0. This is where I get stuck. I know at some point you need to use logs and I think before that you might substitute a letter for a piece of the equation. I am not sure , please help.

Hi, welcome! You've got the right ideas. Substitute \(\displaystyle \L u = 2^x\) to get \(\displaystyle \L u^2 -12u - 1=0\) and use the quadratic equation to solve for \(\displaystyle \L u.\) Then use logs to solve for \(\displaystyle \L x.\) Only the positive solution for \(\displaystyle \L u\) is valid.

 

[jbizhop]

thank you very much, very appreciated