1. ## Exponents and Logarithms

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

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

2. Take logarithms to the base 5 on both sides. You have $\displaystyle 2x = 1-x$ which is a simple linear equation in one variable.

3. For the second one, use the substitution:

$\displaystyle y = 2^{2x}$

4. 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

5. Right, now, replace y by 2^2x again

6. 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?

7. 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?

8. 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.

9. Yes this had really helped a lot.....Question solved.