Can Someone Help Me Solve These For Calculus?

Question

These 3:

3^x+1=5

2 log3x = log325

Solve For X:

2^2x + 2^x = 12

Answer 1

1)

3^x + 1 = 5
3^x = 4
log 3^x = log 4

x log3 = log 4

x = log4/log3 ☺ ANSWER

2)
2 log3x = log325

log (3x)^2 = log 325

9x^2 = 325

x^2 = ­325÷9

x = sqrt (325) / 3 ☺ ANSWER

3)

2^2x + 2^x = 12
2^2x + 2^x -12 = 0 ......................if 2^x = a
a² + a -12 = 0
(a+4)(a-3) = 0
a = -4........cannot be negative
a = 3......ok

2^x = 3

x log 2 = log 3

x = log 3/ log 2 ☺ ANSWER

Answer 2

For the last one, substitute y = 2^x... you now have
y² + y = 12
→ y² + y - 12 = 0
→ (y+4)(y-3) = 0
→ y=-4 or y=3
→ 2^x = -4 or 2^x = 3
→ 2^x = 3 (2^x can't be negative)
→ x = log(base 2) 3 = log3 / log2.

Answer 3

Question 1
3^(x) = 4
x.log 3 = log 4
x = log 4 / log 3
x = 1.26

Question 2
2 log x = log 25
log x = (1/2). log 25
log x = log 25^(1/2)
log x = log 5
x = 5

Question 3
let y = 2^(x)
y² + y - 12 = 0
(y + 4).(y - 3) = 0
y = - 4, y = 3
Taking +ve value for y:-
3 = 2^(x)
log 3 = x.log 2
x = log 3 / log 2
x = 1.58