Question

solve the following equations:
log3 (x+1) = log9 (1-X)

Answer 1

use change of base formulas

Answer 2

cross multiply and hope for the best :)

Answer 3

log {9} (1-x) = (1/2) log {3} (1-x)

Answer 4

where the number in {} is the base

Answer 5

yeah, thats prolly base 3 and base 9 there eh

Answer 6

then you can use normal log laws to collect the logs into a single log

Answer 7

let y= \[\log_{9} (1-x) \]

Answer 8

let y= \[\log_{9} (1-x) \]

Answer 9

so \[9^{y} = 1-x \]

Answer 10

\[3^{2y} = 1-x \]

Answer 11

take log base 3 of both sides , so it matches with the other log in the question

Answer 12

then divide by two , thats how you change the bases

Answer 13

rest is easy

Answer 14

think the only solution is x = 0

this equation says
\[(x+1)^2=1-x\]

lets try this:
\[log_3(x+1)=log_9(1-x)=\frac{log_3(1-x)}{log_3(9)}=\frac{log_3(1-x)}{2}\]
\[2log_3(x+1)=log_3(1-x)\]
\[log_3(x+1)^2=log_3(1-x)\]
\[(x+1)^2=1-x\]
\[x^2+2x+1=1-x\]
\[x^2+3x=0\]
\[x(x+3)=0\]
\[x=0\]or
\[x=-3\] and of course -3 is out because you cannot take the log of a negative number

Answer 15

thank you soo much! i like the way you explaine things