1.Solve, 3^2x = 5^(1-x)

2. Solve 4^2x - 8(2^2x) - 9 =0

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- Oct 30th 2010, 09:46 PMnav92Exponents and Logarithms
1.Solve, 3^2x = 5^(1-x)

2. Solve 4^2x - 8(2^2x) - 9 =0 - Oct 30th 2010, 10:18 PMbandedkrait
Take logarithms to the base 5 on both sides. You have $\displaystyle 2x = 1-x$ which is a simple linear equation in one variable.

- Oct 30th 2010, 10:36 PMUnknown008
For the second one, use the substitution:

$\displaystyle y = 2^{2x}$ - Oct 30th 2010, 10:55 PMnav92
For the second one,

The answer must be in 3 significant numbers which I can't seem to get......Instead I get,

After substitution,

y^2 - 8y - 9 = 0

y= 9 ; y= -1 - Oct 30th 2010, 10:57 PMUnknown008
Right, now, replace y by 2^2x again (Smile)

- Oct 30th 2010, 11:31 PMnav92
Oh man , how could i missed the substitution part on question 2...okay i get it:)

What about question 1 where the base number is different? - Oct 30th 2010, 11:35 PMUnknown008
Hm... let's do it.

$\displaystyle 3^{2x} = 5^{1-x}$

Take logs on both sides.

$\displaystyle log3^{2x} = log5^{1-x}$

$\displaystyle 2x log3 = (1-x)log5$

$\displaystyle 2x log3 = log5 - xlog5$

Bring xlog5 to the left;

$\displaystyle 2x log3 + x log 5 = log5$

Factor x;

$\displaystyle x (2log3 + log 5) = log5$

Can you continue now? (Happy) - Oct 30th 2010, 11:39 PMearboth
- Oct 30th 2010, 11:44 PMnav92
Yes this had really helped a lot...:D..Question solved.