help with tricky logarithm problem? log(base 8) a + log(base 4) b^2 = 5.......?

Question

Determine the value of ab if:

log(base 8) a + log(base 4) b^2 = 5

and

log(base 8) b + log(base 4) a^2 = 7

Answer 1

Paul was right...I did this wrong the first time.

log8 x = log2 x/log2 8 = 1/3 log2 x, and

log4 x = log2 x/log2 4 = 1/2 log2 x

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log8 a + log4 b² = 5

convert to log2:

1/3 log2 a + 1/2 log2 b² = 5

combine:

log2 a^(1/3)b = 5

Remove logs:

a^(1/3)b = 2^5 = 32

a^(1/3)b = 32

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log8 b + log4 a² = 7

convert to base 2:

1/3log2 b + 1/2log2 a² = 7

combine:

log2 b^(1/3)a = 7

Remove logs:

b^(1/3)a = 2^7 = 128

b^(1/3)a = 128

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Now you're looking for ab, so multiply both of the results together:

a^(1/3)b(b^(1/3)a) = 32(128)

a^(4/3)b^(4/3) = 4096

ab = 4096^(3/4) = 8³ = 512

OK...I think that's finally it...

Answer 2

The fact to remember is that

log_b (x) = log_k (x) / log_k (b)

It turns out to be easiest to convert everything to logs in base 2.
2^2 = 4 and 2^3 = 8, which means that
log_2 (4) = 2 and log_2 (8) = 3.

Using the above formula,

log_8 (x) = log_2 (x) / log_2 (8)
log_4 (x) = log_2 (x) / log_2 (4)

You also need to remember that
log_x (y^n) = n * log_x (y)

The first equation therefore becomes
log_2 (a) / 3 + 2 * log_2 (b) / 2= 5

If we set A = log_2 (a) and B = log_2 (b)
then
A / 3 + B = 5

I'm not going to finish your homework for you, but you can do exactly the same thing for the second equation.

This should give two simultaneous equations for A and B. Solve these, in the usual manner. You will then get a and b via
a = 2^A b = 2^B

Answer 3

a million. Log base [5] of (8/12) so Log base 5 of two/3 reason is once you subtract to logarithms of a similar base, you could divide what's in the parentheses. 2. similar challenge because the first one, both logs are base e. ln 16/4 or ln 4 3. log (e4da3b7fbbce2345d7772b674a318d5x^2) 4. 4 log base 16 of 6