can anyone please help with this question log (3x) = log 4 + log (x - 1)
M math321 New member Joined Nov 18, 2010 Messages 9 Nov 18, 2010 #1 can anyone please help with this question Code: log (3x) = log 4 + log (x - 1)
pka Elite Member Joined Jan 29, 2005 Messages 11,995 Nov 18, 2010 #2 That is equivalent to log(4x−43x)=0\displaystyle \log \left( {\frac{{4x - 4}}{{3x}}} \right) = 0log(3x4x−4)=0.
That is equivalent to log(4x−43x)=0\displaystyle \log \left( {\frac{{4x - 4}}{{3x}}} \right) = 0log(3x4x−4)=0.
M math321 New member Joined Nov 18, 2010 Messages 9 Nov 18, 2010 #3 and then wat can u jus break it down a little simpler for me please thank you
L lookagain Elite Member Joined Aug 22, 2010 Messages 3,255 Nov 18, 2010 #4 pka said: That is equivalent to log(4x−43x)=0\displaystyle \log \left( {\frac{{4x - 4}}{{3x}}} \right) = 0log(3x4x−4)=0. Click to expand... math321, this is equivalent to \(\displaystyle \log_{10} \left({\frac{{4x - 4}}{3x}}} \right) = 0\) Change this equation to its equivalent exponential form: 100 = 4x−43x\displaystyle 10^0 \ = \ \frac{4x - 4}{3x}100 = 3x4x−4 1 = 4x−43x\displaystyle 1 \ = \ \frac{4x - 4}{3x}1 = 3x4x−4 Solve for x,\displaystyle x,x, but you must check those candidate solutions in the original equation.
pka said: That is equivalent to log(4x−43x)=0\displaystyle \log \left( {\frac{{4x - 4}}{{3x}}} \right) = 0log(3x4x−4)=0. Click to expand... math321, this is equivalent to \(\displaystyle \log_{10} \left({\frac{{4x - 4}}{3x}}} \right) = 0\) Change this equation to its equivalent exponential form: 100 = 4x−43x\displaystyle 10^0 \ = \ \frac{4x - 4}{3x}100 = 3x4x−4 1 = 4x−43x\displaystyle 1 \ = \ \frac{4x - 4}{3x}1 = 3x4x−4 Solve for x,\displaystyle x,x, but you must check those candidate solutions in the original equation.