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Solve each equation

0 votes

Solve each equation

asked Oct 26, 2018 in ALGEBRA 2 by anonymous

1 Answer

0 votes
1)
 
log 5x  =  log (2x + 9)
 
Apply antilogarithm to each side
 
5x  =  2x + 9
 
5x - 2x  =  9
 
3x  =  9
 
x  =  9/3
 
x  =  3
 
2)
 
log (10 - 4x)  =  log (10 - 3x)
 
Apply antilogarithm to each side
 
(10 - 3x) - (10 - 4x)  =  0
 
10 - 3x - 10 + 4x  =  0
 
x  =  0
 
3)
 
log (4p - 2)  =  log (-5p + 5)
 
Apply antilogarithm to each side
 
4p - 2  =  -5p + 5
 
4p + 5p  =  5 + 2
 
9p  =  7
 
p  =  7/9
 
4)
 
log (4k - 5)  =  log (2k - 1)
 
Apply antilogarithm to each side
 
4k - 5  =  2k - 1
 
4k - 2k  =  -1 + 5
 
2k  =  4
 
k  =  4/2
 
k  =  2
 
5)
 
log(-2a + 9)  =  log (7 - 4a)
 
Apply antilogarithm to each side
 
(-2a + 9)  =  (7 - 4a)
 
-2a + 4a  =  7 - 9
 
2a  =  -2
 
a  =  -2/2
 
a  =  - 1
 
6)
 
2log_7 (-2r)  =  0
 
log_7 (-2r)  =  0
 
Apply logarithmic rule
 
-2r  =  7^0
 
-2r  =  1
 
r  =  - 1/2
 
7)
 
-10 + log_3 (n + 3)  =  - 10
 
-10 + log_3 (n + 3) + 10  =  0
 
log_3 (n + 3)  =  0
 
Apply logarithmic rule
 
n + 3  =  3^0
 
n + 3  =  1
 
n  =  1 - 3
 
n  =  - 2
 
8)
 
-2log_5 (7x)  =  2
 
log_5 (7x)  =  2/(-2)
 
log_5 (7x)  =  - 1
 
Apply logarithmic rule
 
7x  =  5^(-1)
 
7x  =  1/5
 
x  =  1/(7X5)
 
x  =  1/35
answered Oct 29, 2018 by homeworkhelp Mentor

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