Logaritum questions! help. test tomorrow.?

Question

can someone please tell me step by step how to get the answer


X
7 = 3 (of course the "x" is small)


AND this one


log6 4+log6 9 (the "6's" are small.

AND

f(x)=x squared-kx+1. If f(2)=3...then k=?

Answer 1

Might I suggest you learn to spell it first? "Logarithm" is correct.

Answer 2

7^x = 3
x ln 7 = ln 3
x = (ln 3/ln 7)
x = .565

log6 4 + log6 9 = log6 (4 x 9) = log6 36
log6 36 is the same as 6^x = 36
Take the natural log of both sides:
x ln 6 = ln 36
x = ln 36/ln 6 = ln (6^2)/ln 6 = 2 ln 6/ln6 = 2

f(2) means that x = 2 in the function "f"
Therefore, if f(2) = 3, then x^2 - kx + 1 = 3 when x = 2
So, 2^2 - k (2) + 1 = 3
4-2k+1=3
-2k = -2
k = 1

Answer 3

Not sure what you mean by the first one.

In general: log(base N)X can be written as log(X)/log(N) where log(X) and log(N) are to the base 10. This is usually a good first step. You should get at least a little credit even if you don't know what to do next.

log6 4+log6 9 (the "6's" are small.

log(4)/log(6) + log(9)/log(6)
(log(4) + log(9))/log(6) = log(4 * 9)/log(6)
log(36)/log(6) = log(6^2)/log(6) = 2log(6)/log(6) = 2

AND:
f(2) = 3 = 2^2 - k2 + 1
2 = 4 - 2k
2k = 4
k = 2