Question

7log9(x+8)=7

Answer 1

Isolate the log.

You could divide both sides by 7.

Answer 2

that's it? just isolating log and taking it from there?

Answer 3

\[7\log9(x+8)=7\]\[\frac{7\log9(x+8)}{\color{red} {7} }=\frac{7}{\color{red} {7} }\]\[\log9(x+8)=1\]\[\color{green} {Can~~~~you~~~~do~~~~\it~~~~now?}\]

Answer 4

I have up to that part.
log9(x+8)=1, do you move the x+8 to the right or how do you do it?

Answer 5

is 9 the base?

Answer 6

yes

Answer 7

x=1?

Answer 8

just making sure, ok, it's \[\log_9(x+8)=1\]
\[\color{red} {Log_ab=c}~~~~~->~~~~~\color{red} {a^c=b}\]\[9^1=x+8~~~~~~~~~~~~~~~\color{red} {x=?}\]

Answer 9

YEP!!!!

Answer 10

x=1 is correct, but do you know how to solve it if it was more complex? You have to put both sides to the power of 9, or use solomon's method.

Answer 11

teenagerunaway, good work!

Answer 12

Thanks SolomonZelman, i need help on a more complex one, any one up for it?

Answer 13

Yes, you can make a new question and ask it there.

Answer 14

Alright, log 2(a^2-6a)=log2(10+3a)
would the logs cancel out?

Answer 15

\[\log_2(a^2-6a)=\log_2(10+3a)\] THIS?

Answer 16

yes

Answer 17

Yes, logs are one-to-one functions, you can drop both logs here.

Answer 18

if so it becomes,
a^2 -6a=10+3a

Answer 19

Now you for sure can do it.

Answer 20

i ended up with a^2 -9a=10, is that wrong?

Answer 21

now it's not, it is correct, and what do you prefer, quadratic formula, factoring or completing the square?

Answer 22

factoring,
(a-10)(a+1)

Answer 23

(a-10)(a+1) = 0

now solve for a.

Answer 24

Make sure to CHECK your solutions... in the original equation \[\large \log_2(a^2-6a)=\log_2(10+3a)\]

Answer 25

log2(-5)=log2(7)
log2(40)=log2(40)

Answer 26

So...?

Answer 27

see a-10 only works, not a=-1

Answer 28

a=10 is the answer

Answer 29

Correct.

Answer 30

YES!!!!

Answer 31

Thank you so much!!!!!