similar logarithm problem!

Sep 2008
5,237
1,625
Melbourne
\(\displaystyle e^{2x}-7e^x+10=0\)

\(\displaystyle (e^{x})^2-7e^x+10=0\)

Lets make \(\displaystyle e^{x} = a\)

Now we have

\(\displaystyle a^2-7a+10=0\)

which can be factored as follows

\(\displaystyle (a-5)(a-2)=0\)

by the null factor law

\(\displaystyle a=2,5\)

and bring our original unknown back into the picture

\(\displaystyle e^x=2,5\)

\(\displaystyle x=\ln(2),\ln(5)\)

Does this make sense?
 
Jan 2006
118
1
okay, that makes a lot of sense so there are two answers technically and I forgot to take the ln of the 2 and 5. that is where i was going wrong..thanks so much!