Logarithms

mulciber

Hi. I'm a leaving cert student and I'm doing logarithms at the moment. This may be a stupid question but how do you solve equations like this: e^x=2 and ln x=1.

Do you do it like this?: e^x=2, xloge in the base of e= 2, x=2.

And lnx=1, logx to the base of e=log1 to the base of e, x=1.

Any help would be appreciated.

Thanks.

~Mulciber

MathsManiac

The natural log function "ln()" and the exponential function "e^()" are inverse functions of each other. That is, either of them will "undo" the other.

Remember that when you're solving an equation, you need to do the same thing to both sides.

In your first equation, if you take the natural log of both sides, you get ln(e^x) = ln(2).

Because ln() and e^() are inverse functions, the left hand side is just x. So you have your answer: x = ln(2).

You can similarly solve the second one by applying e^() to both sides.

mulciber

MathsManiac wrote: »

The natural log function "ln()" and the exponential function "e^()" are inverse functions of each other. That is, either of them will "undo" the other.

Remember that when you're solving an equation, you need to do the same thing to both sides.

In your first equation, if you take the natural log of both sides, you get ln(e^x) = ln(2).

Because ln() and e^() are incverse functions, the left hand side is just x. So you have your answer: x = ln(2).

You can similarly solve the second one by applying e^() to both sides.

Ok, I understand. Thanks a million.

FoxT

just one other thing....

e^x=2, xloge in the base of e= 2, x=2.

you are correct when you state that ln(e^x) = xloge.

but ln(e) =1 , so xloge = x.

In the above, your mistake was no to take the log of both sides.

Lets go back to this, though:

ln(e^x) = xloge

Are you happy that this is true always?
What if x=0?
What if x is <0?

Think this through, I'm not trying to confuse you, just hope it will help your understanding of it.

-FoxT