ATAR Notes: Forum
VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: Mulan on May 09, 2010, 08:15:21 pm
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how do you solve 5^2x - 5^x+1 = -4 ??
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Let A = 5^x
A^2 - 5A + 4 = 0
use quadratic formula. :)
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Let A = 5^x
A^2 - 5A + 5 = 0
use quadratic formula. :)
thats what I did but i got the wrong answer D:
the answers x = 0.86, 0 but i got x = 1,4
i just factorised it to find x, not the quadratic formula. is that wrong?
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You are almost there,
, thus solve
and
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You are almost there,
, thus solve
and 
omg yea! woopsies.
i forgot that I made 5^x=Y
thanks so much :)
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can you help me with another question? xD
solve 5(3^x) - 40 (3^-x)=17 for x
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multiply both sides by
, notice that you get a quadratic similair to the one before.
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multiply both sides by
, notice that you get a quadratic similair to the one before.
y do you multiply both sides by 3^x?
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So that
you get a quadratic similair to the one before.
It is much easier to solve.

=3^x\cdot 17)
. where
(which holds for all real numbers.)
Then solve as a quadratic in
as before.
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So that you get a quadratic similair to the one before.
It is much easier to solve.

=3^x\cdot 17)
. where
(which holds for all real numbers.)
Then solve as a quadratic in
as before.
ohhhh thanks you guys :) really helped a lot