SOLUTION: could you help me about this?:) log(3,log(4,log(5,x)))=0

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: could you help me about this?:) log(3,log(4,log(5,x)))=0      Log On


   



Question 202703: could you help me about this?:)
log(3,log(4,log(5,x)))=0

Answer by Edwin McCravy(20062) About Me  (Show Source):
You can put this solution on YOUR website!
log%283%2C%28log%284%2C%28log%285%2Cx%29%29%29%29%29=0

We use this principle:

The logarithmic equation log%28B%2CA%29=C may be written
as the exponential equation B%5EC=A

So let's let A=log%284%2C%28log%285%2Cx%29%29%29=0

Then

log%283%2C%28log%284%2C%28log%285%2Cx%29%29%29%29%29=0

becomes the log equation:

log%283%2CA%29=0

which by the above principle (in red) may be written as the
exponential equation:

3%5E0=A or A=1

And since we let A=+log%284%2C%28log%285%2Cx%29%29%29, then

log%284%2C%28log%285%2Cx%29%29%29=1

This time we let A=log%285%2Cx%29, and
that becomes:

log%284%2CA%29=1

Then we use the above same principle in red again on that
logarithmic equation, and we get the exponential equation:

4%5E1=A or A=4

Since we let A=log%285%2Cx%29, 

4=log%285%2Cx%29

then we use that same principle again on that logarithmic
equation, and we get the exponential equation

5%5E4=x

thus

x+=+625

Edwin