MI 3
Log Quiz #2
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Solve the equation for x. Leave answer in terms of logs, simplified as much as possible.
1.
a.
3x  5
b. (1.025)2 x1  5
1
 log1.025 (5)  1
2
(1.025) 2 x 1  5  log1.025 5  2 x  1  x 
3x  5  log 3 5  x
2. Use the Laws of Logarithms to simplify the expression:
 x2  y 2 
 ( x  y)( x  y) 
log5 ( x  y )  log5 ( x  y)  log5 
  log5 
  log5  x  y 
x y


 x y 
2
3.
2
Given p  log3 (2) , q  log3 (5) - Express the follow in terms of p, q if possible.
 25 
a. log 3    log 3 (52 )  log 3 (2)  2q  p
 2 
b.
1
 8 
 8 2
log 3 
  log 3  
 45 
 45 
1
 log 3 (23 )  log 3 (32  5) 
2
1
 3 p  log 3 (32 )  log 3 (5) 
2
1
 3 p  2  q 
2
4.. Solve for x:
a. log 2 ( x)  log 2 (2 x  3)  1 
b. log 2 ( x)  log( x 2 )  15 Let u  log x
log 2 ( x  (2 x  3))  1  2 x 2  3 x  2 
(2 x  1)( x  2)  0  x 
1
or  2
2
Since x  2 , is not in the domain of log 2 ( x), x 
IMSA
Then u 2  2u  15  0  (u  3)(u  5)  0 .
So, u  3 or  5  x  103 or 105
1
2
Sp12
5. Graph each of the following functions. In each case label the scales on the x- and y- axes clearly and
label at least three points.
f ( x)  log3 x
f ( x)  log3 ( x  4)  2
Domain =
Range =
Domain =
Range =
f ( x)  log3  3x   1
Domain =
Range =
f ( x)  log3 x  3
Domain =
Range =
1
x
6. Prove: log b     log b ( x)
IMSA
SP12