problem = solve and give EXACT answer 2^4x = 3^3x+1
F flew62999 New member Joined Aug 30, 2005 Messages 47 Dec 9, 2005 #1 problem = solve and give EXACT answer 2^4x = 3^3x+1
pka Elite Member Joined Jan 29, 2005 Messages 11,971 Dec 9, 2005 #2 \(\displaystyle \L {\rm{2}}^{{\rm{4x}}} = 3^{3x + 1} \; \Rightarrow \left( {4x} \right)\ln (2) = \left( {3x + 1} \right)\ln (3)\)] Now solve for x.
\(\displaystyle \L {\rm{2}}^{{\rm{4x}}} = 3^{3x + 1} \; \Rightarrow \left( {4x} \right)\ln (2) = \left( {3x + 1} \right)\ln (3)\)] Now solve for x.
F flew62999 New member Joined Aug 30, 2005 Messages 47 Dec 9, 2005 #3 now I am really confused...how did you get that? and where did the 1n come from?
pka Elite Member Joined Jan 29, 2005 Messages 11,971 Dec 9, 2005 #4 You could use base 10 log \(\displaystyle \L {\rm{2}}^{{\rm{4x}}} = 3^{3x + 1} \; \Rightarrow \left( {4x} \right)\log (2) = \left( {3x + 1} \right)\log (3)\)
You could use base 10 log \(\displaystyle \L {\rm{2}}^{{\rm{4x}}} = 3^{3x + 1} \; \Rightarrow \left( {4x} \right)\log (2) = \left( {3x + 1} \right)\log (3)\)
stapel Super Moderator Staff member Joined Feb 4, 2004 Messages 16,583 Dec 9, 2005 #5 flew62999 said: now I am really confused...how did you get that? and where did the 1n come from? Click to expand... Do you not know what logarithms are...? (You didn't know that the "ln" was an "ell-enn", for "natural log", is why I ask.) Eliz.
flew62999 said: now I am really confused...how did you get that? and where did the 1n come from? Click to expand... Do you not know what logarithms are...? (You didn't know that the "ln" was an "ell-enn", for "natural log", is why I ask.) Eliz.
U Unco Senior Member Joined Jul 21, 2005 Messages 1,134 Dec 9, 2005 #6 Don't feel bad about not being familiar with "ln", Flew. I thought it was "In" after I could do questions like these (using "logs").
Don't feel bad about not being familiar with "ln", Flew. I thought it was "In" after I could do questions like these (using "logs").
