Solving just these three logarithms?

Question

logx9= -2

log2x+5=8-log2(x+7)

log5x+3=log5(x-20)+4

Answer 1

Note:

log (base a) b = (log b) / (log a)

a log b = log (b^a)

log a + log b = log (ab)

log a - log b = log (a/b)

1)

log (base x) 9 = -2

(log 9) / (log x) = -2

log 9 = -2 log x

log 9 = log (x^-2)

9 = x^-2 = 1/(x^2)

x^2 = 1/9

x = 1/3

2)

log (base 2) x + 5 = 8 - log (base 2) (x+7)

(log x) / log 2 + 5 = 8 - [log (x+7)] / log 2

(log x) / log 2 + 5 = 8 - [log (x+7)] / log 2

[log x + log (x+7)] / log 2 = 8-5

[log x(x+7)] / log 2 = 3

log x(x+7) = 3 log 2 = log (2^3)

x(x+7) = 2^3 = 8

x^2 +7x - 8 = 0

(x+8)(x-1) = 0

x = -8 or 1

3)

log (base5) x+3=log (base5) (x-20) + 4

(log x) / (log 5) + 3 = [log (x-20)] / log 5 + 4

[log x - log (x-20)] / log 5 = 4 -3 = 1

log [x/(x-20)] = log 5

x/(x-20) = 5

x = 5(x-20)

x = 5x - 100

x = 25