M mathxyz Junior Member Joined Jul 8, 2005 Messages 112 Jul 8, 2005 #1 Express as a single log. 1) 8log_2 (sqrt{3x-2}) - log_2 (4/x) + log_2 (4) 2) 2(log_a (5x^3) - 1/2 log_a (2x + 3)
Express as a single log. 1) 8log_2 (sqrt{3x-2}) - log_2 (4/x) + log_2 (4) 2) 2(log_a (5x^3) - 1/2 log_a (2x + 3)
P pav New member Joined Jun 30, 2005 Messages 29 Jul 8, 2005 #3 loga-logb= log (a/b) loga-logb+logc log(a/b)*c
M mathxyz Junior Member Joined Jul 8, 2005 Messages 112 Jul 8, 2005 #4 pav Thank you so much for your help. I have a test on Monday. Jack
M Matt Junior Member Joined Jul 3, 2005 Messages 183 Jul 8, 2005 #6 Depends on how fast you say it. :wink:
M mathxyz Junior Member Joined Jul 8, 2005 Messages 112 Jul 8, 2005 #7 No help I will wait for someone else to reply with my log questions.
M mathxyz Junior Member Joined Jul 8, 2005 Messages 112 Jul 8, 2005 #8 OKAY MY WORK WITH QUESTION 2: Is it right? Express as a single log. 1) 8log_2 (sqrt{3x-2}) - log_2 (4/x) + log_2 (4) 2) 2(log_a (5x^3) - 1/2 log_a (2x + 3) QUESTION 2: log_a (5x^3)^2) - log_a (2x + 3)^1/2 log_a (5x^3)^2 - log_asqrt{2x + 3} I now apply the difference of log rules. My final answer for question 2: log_a (5x^6/sqrt{2x + 3} ) Is it right? Who can help me with question 1 stated above?
OKAY MY WORK WITH QUESTION 2: Is it right? Express as a single log. 1) 8log_2 (sqrt{3x-2}) - log_2 (4/x) + log_2 (4) 2) 2(log_a (5x^3) - 1/2 log_a (2x + 3) QUESTION 2: log_a (5x^3)^2) - log_a (2x + 3)^1/2 log_a (5x^3)^2 - log_asqrt{2x + 3} I now apply the difference of log rules. My final answer for question 2: log_a (5x^6/sqrt{2x + 3} ) Is it right? Who can help me with question 1 stated above?
tkhunny Moderator Staff member Joined Apr 12, 2005 Messages 11,325 Jul 8, 2005 #9 Question #2 - Sooooo Close. That '5' should be a '25'.
G Gene Senior Member Joined Oct 8, 2003 Messages 1,904 Jul 9, 2005 #10 1) 8log_2 (sqrt{3x-2}) - log_2 (4/x) + log_2 (4) = log_2((3x-2)^(1/2)^8) - log_2(4/x)+log_2(4) = log_2((3x-2)^4) - log_2(4/x) + log_2(4) = log_2((3x-2^4)*(x/4)*(4) = log_2((3x-2)^4*x)
1) 8log_2 (sqrt{3x-2}) - log_2 (4/x) + log_2 (4) = log_2((3x-2)^(1/2)^8) - log_2(4/x)+log_2(4) = log_2((3x-2)^4) - log_2(4/x) + log_2(4) = log_2((3x-2^4)*(x/4)*(4) = log_2((3x-2)^4*x)
M mathxyz Junior Member Joined Jul 8, 2005 Messages 112 Jul 9, 2005 #11 Great I did not miss question 2 by much. Gene, thanks for your help with question 1. I get it now.