Solving these logarithms?

Question

logx9= -2

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

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

Answer 1

I'm going to use the underscore _ to denote the base of the logarithm.  Remember that a^b = c is the same as log_a(c) = b.

log_x(9) = -2

x^(-2) = 9

1/x² = 9

1 = 9x²

1/9 = x²

±√(1/9) = x

± 1/3 = x

So x = 1/3 since you want the logarithm to have a positive base.

For the next problem start by getting any term with x on the same side of the equation.

log_2(x) + 5 = 8 - log_2(x + 7)   ...subtract 5 from both sides

log_2(x) = 3 - log_2(x + 7)   ...add log_2(x + 7) to both sides

log_2(x) + log_2(x + 7) = 3

Now remember that log(a) + log(b) = log(ab).

log_2(x(x + 7)) = 3

log_2(x² + 7x) = 3

Use the fact that a^b = c is the same as log_a(c) = b.

2³ = x² + 7x

8 = x² + 7x

0 = x² + 7x - 8

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

x = 1 or x = -8

Since you can't take the logarithm of -8, the answer is x = 1.

log_5(x) + 3 = log_5(x - 20) + 4   ...subtract 4 from both sides

log_5(x) - 1 = log_5(x - 20)   ...subtract log_5(x) from both sides

-1 = log_5(x - 20) - log_5(x)

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

-1 = log_5((x - 20)/x)

-1 = log_5(x/x - 20/x)

-1 = log_5(1 - 20/x)

Use that a^b = c is the same as log_a(c) = b

5^(-1) = 1 - 20/x

1/5 = 1 - 20/x   ...multiply both sides by x

x/5 = x - 20   ...subtract x from both sides

-4x/5 = -20   ...multiply both sides by 5

-4x = -100   ...divide both sides by -4

x = 25

Hope this helps you!