Solve Log(10^(x))=Log(b^(x)) = 1

[nejnadusho]

I have dilema.

Is there any other solution different than 1?


Log(10^(x))=Log(b^(x)) = 1
...10...b

In words (Log(10^(x))with base 10)=(Log(b^(x))with base b) = 1

And is the second function defines the first one?

Legend
^- on power
=- equals

Thank you guys

 

[Gib Z]

Well have you seen the log identity $$\log_a p^m = m \log_a p$$ where a is any real base? Using that for this thing, the equation simplifies to x= x = 1. So really, no, no other solutions lol.

 

[tacman]

for your comment!

Yes, there is another solution to this equation. Since both logs have the same base, we can use the property of logarithms that states: Log(a^b) = b * Log(a). Therefore, we can rewrite the equation as:
x * Log(10) = x * Log(b) = 1
Since Log(10) and Log(b) are constants, we can divide both sides by them:
x = 1/Log(10) = 1/Log(b)
So, the solution to this equation is x = 1/Log(b).
As for your second question, yes, the second function (Log(b^(x)) with base b) defines the first one (Log(10^(x)) with base 10), since they are equal to each other.
I hope this helps clarify your dilemma!