H hope4ever New member Joined Sep 12, 2006 Messages 2 Sep 29, 2006 #1 log5(x-4)= log7x solve for x These are just base 10 logs. log100 = 2 This equation has the same format as log 40 = log (2x20) Since both sides have log base 10, you divide by log base 10 and end up with 5(x-4) = 7x, so 5x -20 =7x x=-10
log5(x-4)= log7x solve for x These are just base 10 logs. log100 = 2 This equation has the same format as log 40 = log (2x20) Since both sides have log base 10, you divide by log base 10 and end up with 5(x-4) = 7x, so 5x -20 =7x x=-10
G galactus Super Moderator Staff member Joined Sep 28, 2005 Messages 7,203 Sep 29, 2006 #2 Do you have?: \(\displaystyle \L\\log[5(x-4)]=log[7x]\) \(\displaystyle \L\\5(x-4)=7x\) \(\displaystyle \L\\5x-20=7x\) \(\displaystyle \L\\-20=2x\) \(\displaystyle \L\\x=-10\) In that event, you're okey-dokey.
Do you have?: \(\displaystyle \L\\log[5(x-4)]=log[7x]\) \(\displaystyle \L\\5(x-4)=7x\) \(\displaystyle \L\\5x-20=7x\) \(\displaystyle \L\\-20=2x\) \(\displaystyle \L\\x=-10\) In that event, you're okey-dokey.
