Help with math - logs,?

Question

Solve:

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

(b)4(x-1) = 7

(c) e^(2x-1) = 8

Answer 1

(a) Since 4 = 2 squared and 8 = 2 cubed then we get

2^[2(x - 1)] = 2^[3(2x+3)]

therefore 2 (x - 1) = 3 (2x + 3)
2x - 2 = 6x + 9
-4x = 11
x = 11/-4

Answer 2

a) remark that 8 = 4*2 and so 8 = 4^1.5
remember (a^n) ^m = a^nm
4^(x-1) = 8^(2x+3) = 4^(1.5*(2x+3))=4^(3x+4.5)
if the powers are the same, the numbers are equal
x-1 =3x+4.5 -2x= 5.5 x= -2.75

b) 4(x-1) =4x-4 =7 or 4x =11 result x=11/4 . It is correct, you check

c)remember e^lnx= x (definition)
2x-1 = ln8 or 2x = ln8+1 =2.08+1=3.08 x=1.54

Answer 3

b) 4(x-1)=7
4x-4=7
4x=11
x=11/4

c) e^(2x-1)=8
(2x-1) ln(e) = ln(8)
2x-1 = ln(8) ------- ln e is 1 (to the base e)
2x=ln(8)+1
x=[ln(8)+1]/2