Solving Logarithmic Equations

Question

Solve for x. Please leave log in your answer.

1. 5 * 11^n-2= 59
2. 9^8x - 10 = 43

Answer 1

(1) There is no x in the equation to solve for. Perhaps you meant to say "solve for n?"

log5 + (n - 2)log11 = log59

(n - 2)log11 = log(59/5)

n = [log(59/5) / log11] + 2

(2)

9^8x = 53

8x log9 = log53

x = log53 / 8log9

Answer 2

the biggest thing to remember with logarithms is that they "undo" exponentiation. Treat exponentiation and logarithms together they way you would treat multiplication and division, or addition and subtraction.

Here are the important rules to remember:

log_a (a^b) = b

a^log_a^(b) = b

those are the only formulas we need for this question.

5*11^n-2 = 59

11^n-2 = 59/5

log₁₁( 11^n-2 ) = log₁₁( 59/5 )

n - 2 = log₁₁( 59/5 )

n = log₁₁( 59/5 ) + 2

if you want to change to log_e (usually written as just log), you can use this "change of base" formula:

log_b(x) = log_a(x) / log_a(b)

so using that eqution

n = log₁₁( 59/5 ) + 2 = (log(59/5) / log(11)) + 2

here is the second part.

9^8x-10 = 43

9^8x = 53

log₉(9^8x) = log₉(53)

8x = log₉(53)

x = log₉(53)/8

x = (log(53)/log(9))/8

x = log(53)/(8*log(9))

(you could simplify this more using the fact that log(9) = log(3²) = 2*log(3), but that isnt so important)

if you are curious, here is the reason why the change of base formula is true:

http://mathforum.org/library/drmath/view/70648.html

hope that helps!