Pre-calculus problem

Question

Use f(x)= log₅(2x-1) to address following problems.

a) Find the domain of f

b) Compute f(1)

c)Find the number a so that f(A)=2

Answer 1

Logarithm functions, no matter what the base, require an argument greater than zero. Example: log₅(0) or log₅(-1) are not defined while log₅(0.1) or log₅(100) are. The reason for this is easy to see if you convert the logarithm into an exponential. log₅(-1) means that I would need to find some number x such that 5^x = -1 and there is no number that would do that.

So, to determine the domain, we just need to say:

(2x - 1) > 0

2x > 1

x > 1/2

so the domain is (0.5, ∞)

f(1) = log₅(2•1 - 1) = log₅(1) so 5 raised to what power equals 1. That would be zero. (5⁰ = 1) so log₅(1) = 0

2 = log₅(2x -1) [converting to an exponential we get:

5² = 2x - 1

25 = 2x - 1

26 = 2x

x = 13

Answer 2

Jadelyn:

log functions are defined for positive values.

a) We need to solve 2x -1 > 0 to determine the domain.

b) I'll leave this part to you.

c) We want to solve log₅(2*A-1) = 2.

Remember log_x(a) = y implies x^y = a

So we need to solve 2*A-1 = 5² =25 . I'll let you figure what A is equal to.