Kahn Problem 1 Solve for x 2 points each a 81x3322 b log2x3l

Kahn Problem 1: Solve for x: (2 points each) a) 81x3-322+ b) log,(2x+3)-log (3x + 1)-

8^(4x - 3) = 32^(2x +1)

Clearly 2^3 = 8 and 2^5 = 32 comes into the fore...

(2^3)^(4x-3) = (2^5)^(2x + 1)

Now, using laws of exponents :
2^(3(4x-3)) = 2^(5(2x+1))

2^(12x - 9) = 2^(10x + 5)

Equating exponents :
12x - 9 = 10x + 5
2x = 14
x = 7

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b)
log 4 (2x + 3) - log 4 (3x + 1) = 1/2

Using law of logs :
log 4 ((2x+3)/(3x+1)) = 1/2

Converting to exponential format :
(2x+3)/(3x+1) = 4^(1/2)

(2x+3)/(3x+1) = 2

Crossmultiply :
2x+3 = 2(3x + 1)

2x + 3 = 6x + 2

4x = 1

x = 1/4

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c)
log(a^3 * sqrtb / c^5)
can be written as :

3log(a) + 1/2log(b) - 5log(c) when expanded

Plug back :
3(4) + 1/2(12) - 5(-8)

12 + 6 + 40

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d)
A = 5000 dollars
t = 4
r = 8% or 0.08

So, we have
A = Pe^(rt)

5000 = P * e^(0.08 * 4)

5000 = Pe^(0.32)

P = 5000/e^0.32

P = 3630.75 dollars