# Exponents and Logarithms

• Oct 30th 2010, 09:46 PM
nav92
Exponents and Logarithms
1.Solve, 3^2x = 5^(1-x)

2. Solve 4^2x - 8(2^2x) - 9 =0
• Oct 30th 2010, 10:18 PM
bandedkrait
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 PM
Unknown008
For the second one, use the substitution:

$\displaystyle y = 2^{2x}$
• Oct 30th 2010, 10:55 PM
nav92
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 PM
Unknown008
Right, now, replace y by 2^2x again (Smile)
• Oct 30th 2010, 11:31 PM
nav92
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 PM
Unknown008
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 PM
earboth
Quote:

Originally Posted by nav92
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?

Re-write the equation:

$\displaystyle 3^{2x} = 5^{1-x}~\implies~9^x=\dfrac5{5^x}$

Multiply through by $\displaystyle 5^x$ and you'll get:

$\displaystyle 9^x \cdot 5^x = 5~\implies~(9 \cdot 5)^x=5$

I'll leave the rest for you.
• Oct 30th 2010, 11:44 PM
nav92
Yes this had really helped a lot...:D..Question solved.