Logarithmic equations

[Janet Ward]

I am lost solving logarithmic equations.
here is an example hopefully you can tell me where I'm going wrong
logx + log(x-1) = log(4x)
logx(x-1) = log(4x)
log (x^2 -x) - log(4x) = 0
log (x^2 - 5x) = 0
x^2 -5x = 1 because 10^0=1
x^2-5x-1=0 now I'm not sure what to do.

 

[Deleted member 4993]

I am lost solving logarithmic equations.
here is an example hopefully you can tell me where I'm going wrong
logx + log(x-1) = log(4x)
logx(x-1) = log(4x)
log (x^2 -x) - log(4x) = 0
log (x^2 - 5x) = 0<<<< This step is incorrect.

logx(x-1) = log(4x) ? x(x-1) = 4x ? x[sup:26sz4e1y]2[/sup:26sz4e1y] - 5x = 0 ? x(x-5) = 0

thus

x = 0...............OR................... x-5 = 0 ? x = 5

x^2 -5x = 1 because 10^0=1
x^2-5x-1=0 now I'm not sure what to do.

 

[galactus]

I am lost solving logarithmic equations.
here is an example hopefully you can tell me where I'm going wrong
logx + log(x-1) = log(4x)
logx(x-1) = log(4x)
log (x^2 -x) - log(4x) = 0
log (x^2 - 5x) = 0<------------Here
x^2 -5x = 1 because 10^0=1
x^2-5x-1=0 now I'm not sure what to do.

You're close.

\(\displaystyle log(x(x-1))=log(4x)\)

Now, just solve \(\displaystyle x^{2}-x=4x\Rightarrow x^{2}-5x=0\)

The thing was, \(\displaystyle log(x^{2}-x)-log(4x)\neq log(x^{2}-5x)\)

But, \(\displaystyle log(x^{2}-x)-log(4x)=log(\frac{x^{2}-x}{4x})\)

Also,log(0) is not 1. It is more like negative infinity.

Actually, it is considered undefined.