Question

hiya need some help on logs and e,
a) log base 3 (x+2) +log base 3 (x-1)=2
b) (e^2x)-(5e^x)-14=0
thanks!

Answer 1

use the log rules, if logs are adding each other, then that means you multiply
Multiplication inside the log can be turned into addition outside the log, and vice versa.
http://www.purplemath.com/modules/logrules.htm
\[\log_3 (x+2) +\log_3 (x-1)=2 \]
\[\log_3 ((x+2)(x-1))=2 \]
\[ \log_3(x^2+x-2)=2\]
now, change the logarithm into exponential form so i can be easily solved.

\[3^2=x^2+x-2\]

Answer 2

I tried solving them both but I got the wrong answers,
A) log base 3 (x+2)(x-1)=2
3^2=(x^2)+x-2
9=x^2 +x-2
x^2+x-11=0
then did quad formula
(-1+sqrt 1-4(1)(11))/2
but it turns out the answer is (-1+3sqrt5)/2
and for b)
2xlne-5xlne=ln14
2x-5x=ln14
x=ln14/-3
but the answer is x=ln7

Answer 3

Logarithm Form
\[\log_a b=c\]

Exponential Form
\[a^c=b\]

Answer 4

for b) (e^2x)-(5e^x)-14=0
(e^2x)-(5e^x)=14
i added 14 to both sides so i can the es alone
now let u = e^x
u^2 - 5u = 14
now bring -14 back to the equation
u^2 - 5u - 14 = 0
(u + 2)(u - 7) = 0
two solutions: u = -2 or u = 7
-2 is not correct because e^x cannot be negative
so e^x = 7;
ln (e(x)) = ln(7)
x = ln(7)

Answer 5

thanks!!!

Answer 6

No Problem :)