log4 (6-x)-log2 8=log9 3?

Question

how to do this question?
solve for x.
answer is -122

Answer 1

log[base 4](6 - x) - log[base 2](8) = log[base 9](3)

First off, two of these logs are solveable directly.

Let x = log[base 2](8). Then
2^x = 8 = 2^3, so
x = 3

Let y = log[base 9](3). Then
9^y = 3
(3^2)^y = 3^1
3^(2y) = 3^1, so
2y = 1, y = 1/2

Given that we solved two of those logs, our equation is now

log[base 4](6 - x) - 3 = 1/2

log[base 4](6 - x) = (1/2) + 3

log[base 4](6 - x) = (7/2)

Now, convert this to exponential form.

4^(7/2) = 6 - x
x = 6 - 4^(7/2)

x = 6 - [4^(1/2)]^7

x = 6 - [2]^7
x = 6 - 128
x = -122

Answer 2

loga b is a number which when a is raised to that number, you get b.

so log2 8 is 3, becuase 2^3 = 8.

also we have log9 3 which will equal 1/2, because 9^(1/2) = 3. Note that raising a number to the 1/2 is taking the square root.

So so far we can reduce you equation down to the following:

log4 (6-x) - 3 = .5

If we move our constants to the right side we get:

log4 (6-x) = (7/2)

The above is equivalent to saying:

4^(7/2) = 6 - x

4^(7/2) = 128, since 4^7 = 16384, and the square root of 16384 is 128. So now we have a simple equation:

128 = 6 - x

From here it is easy to see that

x = -122

--charlie

Answer 3

Since logx y = 1/(logy x),

log9 3 = 1/(log3 9) = 1/2

Also log2 8 = 3, so...

log4 (6-x) = 7/2

4^(7/2) = 6-x

2^7 = 6-x

128 = 6-x

x = -122