How do you prove a logarithm equation?

Question

I need to solving this math problem. Can anyone help??
The question: If log [a] b=1/x and log [b] √a=3x^2, show that x=1/6. Remember to show all work. (Note: the log base is contained in the square brackets, as shown: log [base].)

Answer 1

log [a] b = 1/x and log [b] √a = 3x^2

log [b] √a = 3x^2
(log [a] √a) / (log [a] b) = 3x^2
((1/2)(log [a] a)) / (1/x) = 3x^2
x/2 = 3x^2
3x^2 - x/2 = 0
x(3x - 1/2) = 0
x = 0 or 1/6
If x = 0, then 1/x is undefined.
Hence, x = 1/6 is the final solution.

Answer 2

log_a b = 1/x
lob_b (√a) = 3x²

The first thing to note is that log_a b = ln b/ln a. Similarly, log_b (√a) = ln (√a)/ln b. Thus we have:

ln b/ln a = 1/x
ln (√a)/ln b = 3x²

Second, note that by the laws of logarithms, ln (√a) = 1/2 ln a, so we have:

1/2 ln a/ln b = 3x²
ln a/ln b = 6x²

From this we obtain that:

6x² = ln a/ln b = 1/(ln b/ln a) = 1/(1/x) = x

So 6x² = x, thus either x = 0 or x = 1/6. However, x≠0, since otherwise 1/x would not have been defined. Therefore, x must equal 1/6. Q.E.D.

Answer 3

You can use change of variable...

log[a] b = log[c] b / log[c] a

also... log[a] b = 1 / log[b] a
thus
log[b] a = x
log[b] √a = 3x² = 1/2 log[b] a
x = 6x²
but x cannot be 0
thus
x = 1/6



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