math indices problems?

Question

I am having troubles solving this 2 question. Can anyone show me the solution step by step please?

1) Given that y=ax^(b)+10 and that y=26 when x=2 and y=64 when x=3. Find the values of a and b.
2) Solve 4^(x+1)=2-7(2^x)

Answer 1

1)
26 = a*2^b + 10
16 = a*2^b

64 = a*3^b + 10
54 = a*3^b

54/16 = (a*3^b)/(a*2^b)
27/8 = (3/2)^b
b=3 *

16 = a*2^3 = a*8
a=2

* If you can't tell that b=3 from here, take ln of both sides
ln(27/8) = 1.2164= b*ln(3/2) = .4055*b
b = 1.2164/.4055 = 3

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2)
4^(x+1) = 2 - 7*2^x
4^(x+1) +7*2^x - 2 = 0
(4^(x+1) +7*2^x - 2)/4 = 0/4
4^x + 7/4*2^x - 1/2 = 0
(2*x)^2 + 7/4*2^x - 1/2 = 0 *

Now, let y = 2^x
y^2 + 7/4*y -1/2 = 0

Quadratic it up or Factor this bad boy
(y+2)*(y-1/4) = 0
y = -2 or 1/4

If y=-2 then
2^x = -2 which is impossible.

So y = 1/4
2^x = 1/4
x = -2

* 4^x = (2^x)^2 because
4^x = (2*2)^x = (2^x)*(2^x) = (2^x)^2

Answer 2

1.
26 = a2^b + 10
16 = a2^b
similarly
54 = a3^b
take logarithms to base 10 (or any base)
log16 = log(a) + (b)log2
log54 = log(a) + (b)log3

4log2 = log(a) + (b)log2
3log(3) + log2 = log(a) +(b)log3

Eliminate log(a) by multiplying through by -1
4 log2 = log(a) + (b)log2
-3log(3) - log2 = -log(a) -(b)log3 (and add)

3log2 - 3log3 = b(log2 - log3)
3(log2 - log3) = b (log2 - log3)
b = 3

Hence
16 = a 2^3
a = 2

Ans: a= 2 & b = 3