Log problems?

Question

Hi.
I have the following log problems that I can´t solve. I have the answers but I don´t know how to go about getting it. Could someone please walk me through the solution step by step so I can learn how this is done?
1) 2 log base3 (x-1) + log base3 9 = 2 ; (2 times log base 3 (x-1) + log base 3 and then 9 = 2).
2) ln(x+7) - ln(x+2) = ln(x+1)
Thank you!

Answer 1

1) 2 log base3 (x-1) + log base3 9 = 2
log(base 3) (9(x-1)²) = 2
--> 9(x-1)² = 3² =9
(x-1)² = 1
x=2 because (x-1)>0

2) ln(x+7) - ln(x+2) = ln(x+1)
--> (x+7)/(x+2) = x+1
x+7 = x² + 3x + 2
x² + 2x - 5 = 0
let you finishing (care ln(0) or negative numbers do not exists).

Answer 2

The following working is my answer.

1. 2 log base3 (x-1)+ log base3 (9) = 2
log base3 [(x-1)^2 * 9] =2
log base3 [9(x-1)^2] = 2
Therefore, 9(x-1)^2 = 3^2
9(x - 1)^2 = 9
(x-1)^2 = 9/9 =1
x-1 = (+-)√1
x - 1 = 1
x = 1 + 1 =2
Note: - √1 does not exist if you substitute it in the above equation as it is impossible to take the logarithm of a negative number.


2. In (x+7)/(x+2) = In (x+1)
Therefore, (x+7)/(x+2) = x+1
(x+7) = (x+1)*(x+2)
x+7 = x^2 + 2x +x +2
x+7 = x^2 + 3x +2
-x^2 +x -3x + 7 -2 = 0
-x^2 -2x +5 =0
x^2 + 2x -5 = 0
x^2 + 2x = 5
x^2 + 2x + 1 = 5+1
(x + 1 ) ^2 = 6
x + 1 = (+-)√ 6
x = +√ 6 - 1

Note: x = - √6 - 1 does not exist if you substitute it in the above equation since it is impossible to take the logarithm of a negative number.

Answer 3

Question 1
Assume in solution that log means log to base 3:-
2 log(x - 1) + log 9 = 2
log (x - 1)² + log 9 = 2
log [ 9 (x - 1)² ] = 2
9 (x - 1)² = 3²
(x - 1)² = 1
x² - 2x + 1 = 1
x² - 2x = 0
x (x - 2) = 0
x = 0 , x = 2

Question 2
ln [ (x + 7) / (x + 2) ] = ln (x + 1)
(x + 7) / (x + 2) = x + 1
x + 7 = (x + 2) (x + 1)
x + 7 = x² + 3x + 2
x² + 2x - 5 = 0
x = [ - 2 ± √24 ] / 2
x = [ - 2 ± 2√6 ] / 2
x = - 1 ± √6
Accept +ve value:-
x = - 1 + √6

Hope these are the answers you have!

Answer 4

(1) log base3 [(x-1)^2] + log base3 9 = 2
According to log rules... x log y = log [y^x]
and also, logx + log y = log(xy)
so,
log base3 (9.(x-1)^2) = 2
log base3 [9(x^2 - 2x +1)] = 2
log base3 [9x^2 - 18x +9] = 2
taking antilog on both side will give,
9x^2 - 18x +9 = 3^2
9x^2 - 18x +9 = 9
9x^2 - 18x = 0
x^2 - 9x = 0
x^2 = 9x
x = 9 .......... ANS


(2) In (x+7)/(x+2) = In (x+1)
log rules is, lnx / lny = ln(x-y)
Therefore, (x+7)/(x+2) = x+1
(x+7) = (x+1)(x+2)
x+7 = x^2 + 2x +x +2
x+7 = x^2 + 3x +2
-x^2 +x -3x + 7 -2 = 0
-x^2 -2x +5 =0
x^2 + 2x -5 = 0
x^2 + 2x = 5
x^2 + 2x + 1 = 5+1
(x + 1 ) ^2 = 6
x + 1 = + sq root6 OR x + 1 = - sq root6
x = - 1 + sq root 6 OR x = -1 - sq root 6 ...... ANS

hopoe this helps!