Stuck with logarithms

[rozzer123]

Hi all,

I've been trying to solve this question that involves logarithms and I just can't get anywhere with it!

Here is the problem:

9^x - 5・2^x-3/2 = 2^x+1/2 + 2・3^2x+1

I've familiarized myself with some logarithm rules and this is as far as I got...

=> 3^2x - 5・2^x-3/2 = 2^x+1/2 + 2・3^2x-1
=> log(3^2x) - log(5・2^x-3/2) = log(2^x+1/2) + log(2・3^2x-1)

(Am I even allowed to "log" every term, like how I did in the second line?)

=> 2xlog(3) - [log(5)+log(2^x-3/2)] = (x+1/2)log(2) + log(2) + (2x-1)log
=> 2xlog(3) - log(5) - (x-3/2)log(2) = (x+1/2)log(2) + log(2) + (2x-1)log...

At this point, I am not really sure if I am going in the right direction.

Could someone please suggest some tips or advice on how I should tackle this problem? Any help is very much appreciated

 

[Steven G]

Hi all,

I've been trying to solve this question that involves logarithms and I just can't get anywhere with it!

Here is the problem:

9^x - 5・2^x-3/2 = 2^x+1/2 + 2・3^2x+1

I've familiarized myself with some logarithm rules and this is as far as I got...

=> 3^2x - 5・2^x-3/2 = 2^x+1/2 + 2・3^2x-1
=> log(3^2x) - log(5・2^x-3/2) = log(2^x+1/2) + log(2・3^2x-1)

(Am I even allowed to "log" every term, like how I did in the second line?)

=> 2xlog(3) - [log(5)+log(2^x-3/2)] = (x+1/2)log(2) + log(2) + (2x-1)log
=> 2xlog(3) - log(5) - (x-3/2)log(2) = (x+1/2)log(2) + log(2) + (2x-1)log...

At this point, I am not really sure if I am going in the right direction.

Could someone please suggest some tips or advice on how I should tackle this problem? Any help is very much appreciated

If a+b = c+d then log(a+b) = log(c+d). It is not true that If a+b = c+d then log(a)+log(b) = log(c)+log(d). Same holds is a-b = c-d. So the lines above in bold are not equivalent.

Do not use logs at the very start!!! Note that 3^2x+1 =3^2x*3¹=3*3^2x=3*9^x. Try using this technique to combine like terms. Let's see how far you get with this hint.

 

[rozzer123]

If a+b = c+d then log(a+b) = log(c+d). It is not true that If a+b = c+d then log(a)+log(b) = log(c)+log(d). Same holds is a-b = c-d. So the lines above in bold are not equivalent.

Do not use logs at the very start!!! Note that 3^2x+1 =3^2x*3¹=3*3^2x=3*9^x. Try using this technique to combine like terms. Let's see how far you get with this hint.

Thanks Jomo.

So, I applied your technique to the equation and this is where I'm at!

=> 9x - 5・2x・1/√8 = 2x・√2 + 2・3・9x
=> 9x - 5・2x・√8/8 = 2x・√2/2 + 2・3・9x

Am I approaching it correctly?

 

[Steven G]

Hi all,

I've been trying to solve this question that involves logarithms and I just can't get anywhere with it!

Here is the problem:

9^x - 5・2^x-3/2 = 2^x+1/2 + 2・3^2x+1

For the record, your problem does not involve logs at all. Just because you may use logs to solve it I do not feel that it is a log problem.
Is 7-(-2) a subtraction problem or an addition problem?

 

[Steven G]

Thanks Jomo.

So, I applied your technique to the equation and this is where I'm at!

=> 9x - 5・2x・1/√8 = 2x・√2 + 2・3・9x
=> 9x - 5・2x・√8/8 = 2x・√2/2 + 2・3・9x

Am I approaching it correctly?

What happened to the powers??

 

[Otis]

… Here is the problem:

9^x - 5・2^x-3/2 = 2^x+1/2 + 2・3^2x+1 …

Hello rozzer. That equation has no Real solutions. (You didn't post the exercise instructions, so I've assumed that you were asked to solve for x.)

If you've correctly posted the given equation, then your materials probably contain a mistake.

?

 

[JeffM]

In general, logs are not helpful, at least not directly, when dealing with sums and differences of variables.

The first thing I would do is to get powers of like bases gathered.

[MATH]9^x - 5 * 2^{\{x - (3/2)\}} = 2^{\{x+(1/2)\}} + 2 * 3^{(2x+1)} \implies[/MATH]
[MATH]3^{(2x)} - 2 * 3^{(2x+1)} = 2^{\{x+(1/2)\}} + 5 * 2^{\{x - (3/2)\}} \implies[/MATH]
[MATH]3^{(2x)}(1 - 2 * 3^1) = 2^x(2^{1/2} + 5 * 2^{-2}) \implies[/MATH]
[MATH]-\ 5 * 9^x = \dfrac{5 + 4\sqrt{2}}{4} * 2^x.[/MATH]
As Otis says, there is no real solution to that mess. Did you make a mistake somewhere along the way or write the problem down wrong?

 

[Dr.Peterson]

9^x - 5・2^x-3/2 = 2^x+1/2 + 2・3^2x+1

Please check that you copied the problem correctly. I wonder if it might have been something like 9^x - 5・2^(x-3)/2 = 2^(x+1)/2 + 2・3^2x+1. Or it might be a little different.

 

[Steven G]

So (9/2)^x=-(5+4*sqrt(2))/20 so there is no solution.

It seems to me that even if it was copied wrong that the procedure up my equation above would still be the same, hence it still benefits the OP.

IF the rhs was positive now is the time to take logs!

 

[JeffM]

So (9/2)^x=-(5+4*sqrt(2))/20 so there is no solution.

It seems to me that even if it was copied wrong that the procedure up my equation above would still be the same, hence it still benefits the OP.

IF the rhs was positive now is the time to take logs!

I certainly did not mean to imply that your procedure was not beneficial; in fact, I used it although it was not my first step.

My point was entirely different: a student cannot learn a technique from a defective problem. Psychologically, technique becomes solidified in a student's mind by its success in solving a problem. So we needed to change the focus for the time being in another direction. I am sorry if my post was unclear in its intent.

 

[Steven G]

Jeff,
There was no need to make your last post as my post above was not really a reply to your comment. I did however feel that the last (mathematical) line in your post may not have shown the OP why there was no solution so I attempted to show a bit more work. My remark was to the OP that the problem, in my opinion, was not so bad even though it had no solution since they still had to do almost all the work compared to a problem that had a solution. If it did have a solution then there would have been more work (and practice!) for the OP to do.

All is good!