Exponential equations???

Question

A. Simplify:

1. (25^3/2) / (5)^-1/2


B. Solve for x:

1. (0.04)^x+2 = 256

Note: In A, 3/2 is the exponent of 25, while -1/2 is the exponent of 5. In B, x+2 is the exponent of 0.04.







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Answer 1

Question 1
(25)^(3/2) (5^(1/2))
125 x 5^(1/2)
5³ x 5^(1/2)
5^(7/2)

Question 2
(0.04)^(x + 2) = 256
(x + 2) log (0.04) = log 256
(x + 2) = log 256 / log (0.04)
x + 2 = - 1.717
x = - 3.717

NB
log can be to any base in question 2.

Answer 2

Remember the rules for exponents:
a^(-b) = 1 / (a^b)
(ab)^c = (a^c)(b^c)
(a^b)^c = a^(bc)
(a^b) * (a^c) = a^(b+c)
(a^b) / (a^c) = a^(b-c)

So here we have
1)
[ 25^(3/2) ] / (5^(-1/2))
[ 25^(3/2) ] * (5^(1/2))
[ (25^(1/2))^3 ] * (5^(1/2))
[ 5^3 ] * (5^(1/2))
5^(3 + 1/2)
5^(7/2)

For the second one, you can try making the bases match on both sides, but in this case it won't work. You need to use the fact that log(a^b) = b log(a). Take the log of both sides. This gets the (x+2) out of the exponent. Then use the properties of logs to simplify.

Answer 3

A. 25^(7/4)
B. x+ 2 = (log 256)/(log 0.04), so X = (log 256)/(log 0.04) - 2