Question

solve 1) log(x+5)-log(x+1)=log(3x)
2) log5(x-6)=1-log5(x-2)

Answer 1

Do you remember your log rules?

Answer 2

\[\log_b M - \log_b N = \log_b \frac{ M }{ N }\]

Answer 3

based off this rule what can you rewrite the first problem as?

Answer 4

log x+5/x+1 = 3xlog

Answer 5

?

Answer 6

yes but it is log 3x

Answer 7

log = log_10
so
\[\log_{10} \frac{ x+5 }{ x+1 } = \log_{10} 3x\]

Answer 8

oh okayy

Answer 9

let me try again

Answer 10

can you solve it from there?

Answer 11

I believe so :) thanks. what about the 2nd one?

Answer 12

Move the - log to the other side and use the rule:
\[\log_b M + \log_b N = \log_b (M*N)\]

Answer 13

\[\huge \log_5(x-6)=1-\log_5(x-2)\]
\[\huge \log_5 (x-6) + \log_5(x-2) = 1\]

Answer 14

so then would it be x-6+x-2=5? and so on?

Answer 15

(x-6)(x-2) = 5^1

Answer 16

oh okay got it !

Answer 17

thank you so much :)

Answer 18

glad to be of help :)