Question

Can anyone pls help me with this?
Solve the logarithmic equation algebraically. Approximate the result to three decimal places. (If there is no solution, enter NO SOLUTION.)
ln x + ln(x + 7) = 3

Answer 1

@tcarroll010

Answer 2

Try combining the natural logs, then do something to both sides of the equation

Answer 3

In(x(x+7)) = 3

Answer 4

and e^in (x(x+7)) = e^3, right?

Answer 5

yes very good c:

Answer 6

then x^2 + 7x = e^3?

Answer 7

thank you for the medal, @math>philosophy

Answer 8

Yes, good. And since we have this ugly exponential term, this won't end up factoring nicely. We'll have to throw it into the quadractic formula :D

Answer 9

how?

Answer 10

x^2 + 7x - e^3 = 0

This is of the form:
ax^2 + bx + c = 0

Remember how to solve for x? :D

Answer 11

yes i do but what to do w/ the e?

Answer 12

Remember e is just a constant, like pi

Answer 13

You certainly could punch it into your calculator if you wanted to e^3 = 20.08553692...

Or you can leave it in as an exponential til the end. Your answer might turn out more accurately if you don't use the decimal right now :D

Answer 14

ok i will solve it and let u know

Answer 15

is it 2.186?

Answer 16

Hmm I came up with different numbers. Lemme see what went wrong.

Answer 17

\[\large 0=x^2+7x-e^3\]
\[\large x=\frac{-7 \pm \sqrt{49+4e^3}}{2}\]

Answer 18

what did u get?

Answer 19

I got ~2.186 as one of the solutions

Answer 20

yea i got that too

Answer 21

Ok i musta made a mistake :) If you have that same setup for your quadratic formula, then you're probably doing it correctly.

Answer 22

its ok :)

Answer 23

2.186 yah that sounds right.
And of course, you should have 2 solutions, since it's plus/minus the sqrt.

But that value doesn't lie in the domain of our original function since we can't take the natural log of a negative value! :O

You didn't mention that answer, so maybe you already figured that out :D

Answer 24

i got it right, thnx to u an @math>philosophy