Question

solve:
log (x+1) + log (x+4)=1

Answer 1

First combine the two logs via the multiplication rule.

Answer 2

log((x+1)(x+4)) = 1

Answer 3

Then convert to exponent form. (the base is 10 on this one)

Answer 4

so i did X^2+5x+3=0 and stuck after that

Answer 5

+3???

Answer 6

should be -6

Answer 7

\[x ^{2}+5x-6\]

Answer 8

The base is 10

Answer 9

how did you get -6?

Answer 10

so
10 = (x+1)(x+4)

Answer 11

the solutions must be >-1

Answer 12

That make sense?

Answer 13

ohh ok! but what would i do with the 1?

Answer 14

The 1 is your exponent man.

Answer 15

\[10^{1}=(x+1)(x+4)\]

Answer 16

Just simplify that 1 out of the equation...you know.

Answer 17

i wrote it down. i got it. let me try solving it and ill get back to you

Answer 18

Sounds good man.

Answer 19

Don't forget to mark a Best Response and take care...I'll be around for little while longer, if you need something else just holler.

Answer 20

you will find two solutions...one of them is not match !

Answer 21

so i got the solutions x=-6 and x=1 but we can't use negatives in log equations right so the answer would be 1?

Answer 22

yeahh ...true...x must be x>-1 ! so the only solution is 1

Answer 23

Good job man

Answer 24

thanks for all your help!

Answer 25

Anytime. Take care!

Answer 26

you're welcome :) !

Answer 27

can you guys help me with one more problem please? \[3e ^{3x-1}-2=13\]

Answer 28

You got lucky man.
1) Add 2 to both sides.

Answer 29

so i know my first step is to add 2

Answer 30

2) Divide by 3

Answer 31

ok did that

Answer 32

take the natural log (ln) of both sides.

Answer 33

...in order to bring down the exponent on the left.

Answer 34

and eliminates the e

Answer 35

This give:
3x - 1 = ln(5)

Answer 36

ok so would it just be 3x-1=ln5

Answer 37

Add 1. Divide by 3.

Answer 38

Yep.

Answer 39

ok cool! i know where i messed up. Thanks again for your help :)

Answer 40

No prob. Later