Confusion on this logarithmic equation NEED HELP

[Calabi_Yau]

e^(3+2logx) = (3x-2)*e^3

I can get down to this point: e^(logx^2)=3x-2 I have checked the solution in the book, and the steps. Turns out it simplifies to x^2 -3x +2=0. The problem is I can't figure out how I pass from e^(logx^2) to x^2, because as far as I know logx isn't the same as lnx.

Could you please explain me that step?

 

[LCKurtz]

e^(3+2logx) = (3x-2)*e^3

I can get down to this point: e^(logx^2)=3x-2 I have checked the solution in the book, and the steps. Turns out it simplifies to x^2 -3x +2=0. The problem is I can't figure out how I pass from e^(logx^2) to x^2, because as far as I know logx isn't the same as lnx.

Could you please explain me that step?

Apparently log x does mean ln x in the context of that problem.

 

[Calabi_Yau]

Yeah but why?

 

[LCKurtz]

Some texts use the convention that log(x) means natural log if no other base is specified.

 

[symbolipoint]

e^(3+2logx) = (3x-2)*e^3

I can get down to this point: e^(logx^2)=3x-2 I have checked the solution in the book, and the steps. Turns out it simplifies to x^2 -3x +2=0. The problem is I can't figure out how I pass from e^(logx^2) to x^2, because as far as I know logx isn't the same as lnx.

Could you please explain me that step?

I was able to find the same quadratic equation as you after taking natural logarithms of both sides, [STRIKE]but I still could not show that the original left side and right side are equal[/STRIKE] (I was thinking the question wronly. SOLVING the equation was intended, NOT proving left and right were equal).

One way was to take logs of both sides. Another way was to first divide both sides by e^3.
If divide L and R by e^3,
e^(2lnx) = 3x-2
e^(lnx^2)=3x-2
x^2=3x-2
Simple and easy to find x.

 

[HallsofIvy]

In "pre-Calulus" textbooks, and many older "Calculus and higher" text, where using logarithms to do calculations was predominant, "log x" typically meant "common logarithm, base 10" while "ln x" was the "natural logarithm, base e".

However, in modern, Calculus and above", texts, common logarithms are just never mentioned and "log x" is used to mean "natural logarithm".

 

[symbolipoint]

Post #6:
Numbers could be written in "scientific notation", using the base-ten system, and the numbers and tables of logarithms could be used in performing computations for the numbers. This is why base-ten logarithms were important and very often used. Our books taught about both log base ten AND log base e. Best of memory is like HallOfIvy said, log was for base ten, and ln was for base of Euler number.

 

[HallsofIvy]

Post #6:
Numbers could be written in "scientific notation", using the base-ten system, and the numbers and tables of logarithms could be used in performing computations for the numbers. This is why base-ten logarithms were important and very often used. Our books taught about both log base ten AND log base e. Best of memory is like HallOfIvy said, log was for base ten, and ln was for base of Euler number.

However, my point was that most advanced texts simply ignore
"log base 10" and use "log x" to mean natural logarithm.