How would you solve this logerithm problem (precalc?)

1) Given log(base a) of 3 = .5 then log(base a) of 1/9 = ?

I completely forgot precalc and need to take a placement test for college tomorrow...

2) Also how would I solve: If f (x) = x2 − 3x +1 then f (x − 2) ?

log(base a) of 3 = .5
a^.5 = 3
a=9

log(base a) of 1/9 = x
log(base 9) of 1/9 = x
9^x = 1/9
x=-1
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If f (x) = x2 − 3x +1 then f (x − 2)
f(x-2) = (x-2)^2 - 3(x-2) +1
f(x-2) = x^2-4x+4-3x+6+1
f(x-2) = x^2 - 7x + 11

Good luck on your test mate!

Hi,
1) Log(base a) of 1/9 would be Log(base a) 1 - log(base a) 3 squared. Since Log(base a) 1 would be equal to 0 and log(base a) 3 squared can be reduced to 2 x log(base a) 3, the answer would be 0 - 2(0.5) = -1

2) You need to just replace all instances of x with (x-2) in the given function. That would make the function f(x-2) = (x-2)2 - 3(x-2) + 1, which would reduce to f(x-2) = 2x - 4 - 3x + 6 + 1, which would give the answer as f(x-2) = 3-x

All the best for your test!

I use log_a for log (base a)

log_a (3) = .5 use everything as an exponent of a

a^(log_a (3)) = a^.5 a^x and log_2 are inverse functions, and cancel each other out.

3 = a^.5 --> a = 9

so log_a (1/9) = -1

Prob 2
f(x) = x^2 - 3x + 1 then find f (f -2) replace each x with (x - 2)

f(x - 2) = (x- 2)^2 - 3(x -2) + 1 Multiply it all out
(x^2 - 4x + 4) - 3x + 6 + 1 combine like terms

x^2 - 7x + 11

assume base a:


log(3) = .5

This means a^(1/2) = 3

9^(1/2) = 3, and so a = 9.


log(x) = 1/9

by this

9^(x) = 1/9

And then it's obvious what x =

9^(-1) = 1/9

and there you are for the first one.


2)

f(x) = x^2 - 3x + 1

and so what does f(x-2) equal?

Plug it in.

f(x-2) = (x-2)^2 - 3(x-2) + 1

And solve from there. ^^