# Logarithms and exponential functions Watch

Announcements

Report

#2

(Original post by

|2^x+2+1|=|2^x+12| solve giving your ans to 3 sf.

**Shas72**)|2^x+2+1|=|2^x+12| solve giving your ans to 3 sf.

0

reply

(Original post by

How do you think you'll remove the absolute value signs? I'm presuming you'd then be able to solve using logs (or something similar)?

**mqb2766**)How do you think you'll remove the absolute value signs? I'm presuming you'd then be able to solve using logs (or something similar)?

0

reply

**mqb2766**)

How do you think you'll remove the absolute value signs? I'm presuming you'd then be able to solve using logs (or something similar)?

0

reply

Report

#5

(Original post by

I tried to the same way I did with modules that's (2^x+2+1)^2=(2^x+12)^2

**Shas72**)I tried to the same way I did with modules that's (2^x+2+1)^2=(2^x+12)^2

Perhaps a bit of practice about removing the absolute value signs so that they're singled signed on a small number of intervals is easier. Something like for a > b > 0

|x - a| = |x - b|

becomes

x - a = x - b when x > a

-(x - a) = -(x - b) whea x < b

-(x - a) = (x - b) whea a > x > b

0

reply

(Original post by

I got the answer thanks.

**Shas72**)I got the answer thanks.

0

reply

Report

#7

(Original post by

I dont understand how to do this 3^2|x|=6(3^|x|)+16.

**Shas72**)I dont understand how to do this 3^2|x|=6(3^|x|)+16.

0

reply

(Original post by

What have you thought of?

**mqb2766**)What have you thought of?

I dont get the ans

0

reply

Report

#9

0

reply

(Original post by

What have you thought of?

**mqb2766**)What have you thought of?

0

reply

Report

#11

(Original post by

I tried of squaring both sides, I dont get the ans

**Shas72**)I tried of squaring both sides, I dont get the ans

What is the complicated part of this equation, how can you possibly transform it first to make it simpler, then remove the absolute value functions?

0

reply

(Original post by

As said earlier in the thread, squaring isn't always a good method to solve absolute value problems. In this case, this is so.

What is the complicated part of this equation, how can you possibly transform it first to make it simpler, then remove the absolute value functions?

**mqb2766**)As said earlier in the thread, squaring isn't always a good method to solve absolute value problems. In this case, this is so.

What is the complicated part of this equation, how can you possibly transform it first to make it simpler, then remove the absolute value functions?

0

reply

(Original post by

If x>0, then that is correct.

**mqb2766**)If x>0, then that is correct.

0

reply

Report

#15

(Original post by

But the ans is +/_ 1.89

**Shas72**)But the ans is +/_ 1.89

0

reply

(Original post by

Sounds about right. What have you done so far?

**mqb2766**)Sounds about right. What have you done so far?

0

reply

Report

#17

(Original post by

So I did let u=3^x. Then I got quadratic equation. I got u= 8 and u=-2

**Shas72**)So I did let u=3^x. Then I got quadratic equation. I got u= 8 and u=-2

0

reply

Report

#19

(Original post by

X=1.89

**Shas72**)X=1.89

Note, a simple symmetry argument would give the second solution, without solving the other quadratic.

0

reply

(Original post by

Which is 1/2 the answer because you've assumed x>0. Now assume x<0, what does the original equation become, and hence solve it?

Note, a simple symmetry argument would give the second solution, without solving the other quadratic.

**mqb2766**)Which is 1/2 the answer because you've assumed x>0. Now assume x<0, what does the original equation become, and hence solve it?

Note, a simple symmetry argument would give the second solution, without solving the other quadratic.

So I did 3^2x=6.3^x+16 is one equation. What will the other equation be is it 3^2x =-6.3^x -16

0

reply

X

### Quick Reply

Back

to top

to top