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

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

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- Oct 30th 2010, 10:46 PMnav92Exponents and Logarithms
1.Solve, 3^2x = 5^(1-x)

2. Solve 4^2x - 8(2^2x) - 9 =0 - Oct 30th 2010, 11:18 PMbandedkrait
Take logarithms to the base 5 on both sides. You have which is a simple linear equation in one variable.

- Oct 30th 2010, 11:36 PMUnknown008
For the second one, use the substitution:

- Oct 30th 2010, 11:55 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, 11:57 PMUnknown008
Right, now, replace y by 2^2x again (Smile)

- Oct 31st 2010, 12:31 AMnav92
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 31st 2010, 12:35 AMUnknown008
Hm... let's do it.

Take logs on both sides.

Bring xlog5 to the left;

Factor x;

Can you continue now? (Happy) - Oct 31st 2010, 12:39 AMearboth
- Oct 31st 2010, 12:44 AMnav92
Yes this had really helped a lot...:D..Question solved.