Logarithmic and exponential equations

[j9mom]

Homework Statement

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

Homework Equations

Using log or exponential rules

The Attempt at a Solution

First multiply both sides by 3 so 2^x-2^-x=12
I thought I could take the log of both sides then condense the log, but that is not right.

I also attempted to rewrite as 2^x - 1/(2^x) = 12 but I could nowhere with that.

I also attempted to do (2 - 2^-1)^x = 12
(2-1/2)^x = 12
(3/2)^x = 12

log(3/2)^x = log 12

x(log 3/2) = log 12

x= log 12/log 3/2 = 6.128

But that is not the right answer it is supposed to be 3.595 . Where did I go wrong?

 

[Dick]

Homework Statement

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

Homework Equations

Using log or exponential rules

The Attempt at a Solution

First multiply both sides by 3 so 2^x-2^-x=12
I thought I could take the log of both sides then condense the log, but that is not right.

I also attempted to rewrite as 2^x - 1/(2^x) = 12 but I could nowhere with that.

I also attempted to do (2 - 2^-1)^x = 12
(2-1/2)^x = 12
(3/2)^x = 12

log(3/2)^x = log 12

x(log 3/2) = log 12

x= log 12/log 3/2 = 6.128

But that is not the right answer it is supposed to be 3.595 . Where did I go wrong?

You can't do most of those things. 2^x - 1/(2^x) = 12 is a good start. Put u=2^x. Then 2^(-x)=1/u. Form an equation for u, solve for it and then find x.

 

[j9mom]

Ok, so u - 1/u = 12

so u^2 - 12u - 1 = 0

Do I use the quadratic equation to solve for u?

Sorry, but this is unlike any of the other equations we did.

 

[Mentallic]

Ok, so u - 1/u = 12

so u^2 - 12u - 1 = 0

Do I use the quadratic equation to solve for u?

Well, if you were given that quadratic to solve, what would you do?

Sorry, but this is unlike any of the other equations we did.

Then it's probably one of the harder questions that tests your ability to apply other methods you've previously learned to solve the problem.

 

[j9mom]

OK, at first I did not think I should do that, because I would wind up with two answers, but one is negative, which obviously is an extraneous answer because logs cannot be negative. So the other answer was 12.08 so the log 12.08/log 2 = 3.585 which is close to the book's answer.

 

[Mentallic]

OK, at first I did not think I should do that, because I would wind up with two answers, but one is negative, which obviously is an extraneous answer because logs cannot be negative. So the other answer was 12.08 so the log 12.08/log 2 = 3.585 which is close to the book's answer.

Precisely