expert at logarithms?

Question

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Answer 1

I assume the last logx is log to the base 2 of x. Then all logs are base 2 and we have:

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

Exponentiating we have

(x + 2)/4 = 3/log x
log x = 4*3/(x + 2)
log x = 12/(x + 2)

Exponentiating again we have

x = 2^{12/(x + 2)}

By inspection try x = 4

4 = 2^{12/(4 + 2)} = 2^(12/6) = 2^2

This checks so x = 4.

Answer 2

So you want to solve

log[base 2](x + 2) - 2 = log[base 2](3) - log[base 2](logx)

First thing to do is bring the log[base 2](3) to the left hand side, and the -2 to the right hand side.

log[base 2](x + 2) - log[base 2](3) = 2 - log[base 2](logx)

Now, using the log property

log[base b](a/c) = log[base b](a) - log[base b](c), we can combine the left hand side into one log.

log[base 2]( [x + 2]/3 ) = 2 - log[base 2](logx)

Changing this to exponential form (note that log[base b](a) = c if and only if b^c = a), we get

2^(2 - log[base 2](logx)) = (x + 2)/3

We can now decompose the exponential, to:

(2^2) / (2^(log[base 2](logx)) = (x + 2)/3

Multiply both sides by the denominator, to obtain

2^2 = [(x + 2)/3] [2^(log[base 2](logx)]

Note that the 2^ and the log[base 2] are inverses of each other, and thus cancel each other out, leaving us with

2^2 = [(x + 2)/3] [logx]
4 = [(x + 2)/3] [logx]

Multiply both sides by 3 to obtain

12 = (x + 2)logx

We can't solve this using elementary methods.

Answer 3

log[(x+2).logx] =log (3.4)
(x+2)logx=12
logx=12/(x+2)

There may be an extra term of log
Then
log[(x+2)x] =log (3.4)
(x+2)x=12
x^2+2x-12=0
x>0

x= - 1 + sqrt(13)