Logarithms and exponential functions Watch

Shas72
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#1
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#1
|2^x+2+1|=|2^x+12| solve giving your ans to 3 sf.
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mqb2766
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(Original post by Shas72)
|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)?
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Shas72
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(Original post by 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)?
I tried to the same way I did with modules that's (2^x+2+1)^2=(2^x+12)^2
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Shas72
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(Original post by 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)?
I got the answer thanks.
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mqb2766
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(Original post by Shas72)
I tried to the same way I did with modules that's (2^x+2+1)^2=(2^x+12)^2
While its one way to do it, it can make the expression more complex to solve. If you do this, expand the brackets and you have a disguised quadratc in z = 2^x.
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
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Shas72
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(Original post by Shas72)
I got the answer thanks.
I dont understand how to do this 3^2|x|=6(3^|x|)+16.
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mqb2766
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(Original post by Shas72)
I dont understand how to do this 3^2|x|=6(3^|x|)+16.
What have you thought of?
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Shas72
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(Original post by mqb2766)
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
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mqb2766
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(Original post by Shas72)
When I do 3^2x=6.3^x+16 and 3^2x=-6.3^x-16
I dont get the ans
I don't see how you get that. Either remove the absolute value signs in the original equation or transform it first into something simpler. I'd do the latter.
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Shas72
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(Original post by mqb2766)
What have you thought of?
I tried of squaring both sides, I dont get the ans
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mqb2766
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#11
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(Original post by Shas72)
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?
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Shas72
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(Original post by 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?
So is it 3^2x=6(3^x)+16
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mqb2766
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(Original post by Shas72)
So is it 3^2x=6(3^x)+16
If x>0, then that is correct.
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Shas72
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(Original post by mqb2766)
If x>0, then that is correct.
But the ans is +/_ 1.89
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mqb2766
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(Original post by Shas72)
But the ans is +/_ 1.89
Sounds about right. What have you done so far?
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Shas72
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(Original post by mqb2766)
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
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mqb2766
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(Original post by Shas72)
So I did let u=3^x. Then I got quadratic equation. I got u= 8 and u=-2
So what is x?
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Shas72
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(Original post by mqb2766)
So what is x?
X=1.89
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mqb2766
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#19
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(Original post by Shas72)
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.
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Shas72
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(Original post by 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.
Iam still having problem understanding this.
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
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