# Logarithms and exponential functionsWatch

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1 week ago
#21
(Original post by Shas72)
So I use substitution let u=x^5/2
u = x^5/4 rather than x^5/2.
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#22
(Original post by Pangol)
The idea of treating it like a quadratic is that you don't need logs at all. If you put u = x^1.25, then u^2 = x^2.5. What does this turn the original equation into?

(You don't have to do this to solve it without logs, and I think that using them has made this look harder than it is.)
So I get u=5 and u=-4. After this what do I do?
0
1 week ago
#23
(Original post by Shas72)
So I get u=5 and u=-4. After this what do I do?
I think you have factorised incorrectly.

If u = x^1.25, then x^2.5=20x^1.25 becomes u^2 = 20u. What next?
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#24
(Original post by Pangol)
I think you have factorised incorrectly.

If u = x^1.25, then x^2.5=20x^1.25 becomes u^2 = 20u. What next?
Iam sorry. Yeah so I get x=0 and x^1.25=20
0
#25
(Original post by Pangol)
I think you have factorised incorrectly.

If u = x^1.25, then x^2.5=20x^1.25 becomes u^2 = 20u. What next?
Thank you so so much for helping. Thanks
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