Logarithms and exponential functions

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

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)?

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)?

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

I got the answer thanks.

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

What have you thought of?

When I do 3^2x=6.3^x+16 and 3^2x=-6.3^x-16
I dont get the ans

What have you thought of?

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

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?

So is it 3^2x=6(3^x)+16

If x>0, then that is correct.

But the ans is +/_ 1.89

Sounds about right. What have you done so far?

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

So what is x?

X=1.89

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.