can anyone solve this, please? 3^(2x+3)=2^(2-3x) I have been fighting with it for 2 days! Thanks
S Susan Symonds New member Joined Nov 6, 2005 Messages 1 Nov 6, 2005 #1 can anyone solve this, please? 3^(2x+3)=2^(2-3x) I have been fighting with it for 2 days! Thanks
pka Elite Member Joined Jan 29, 2005 Messages 11,995 Nov 6, 2005 #2 32x+3=22−3x\displaystyle {\rm{3}}^{{\rm{2x + 3}}} = 2^{2 - 3x}32x+3=22−3x (2x+3)ln(3)=(2−3x)ln(2)\displaystyle \left( {2x + 3} \right)\ln (3) = \left( {2 - 3x} \right)\ln (2)(2x+3)ln(3)=(2−3x)ln(2) [2ln(3)+3ln(2)]x=2ln(2)−3ln(3)\displaystyle \left[ {2\ln (3) + 3\ln (2)} \right]x = 2\ln (2) - 3\ln (3)[2ln(3)+3ln(2)]x=2ln(2)−3ln(3) Now you solve for x.
32x+3=22−3x\displaystyle {\rm{3}}^{{\rm{2x + 3}}} = 2^{2 - 3x}32x+3=22−3x (2x+3)ln(3)=(2−3x)ln(2)\displaystyle \left( {2x + 3} \right)\ln (3) = \left( {2 - 3x} \right)\ln (2)(2x+3)ln(3)=(2−3x)ln(2) [2ln(3)+3ln(2)]x=2ln(2)−3ln(3)\displaystyle \left[ {2\ln (3) + 3\ln (2)} \right]x = 2\ln (2) - 3\ln (3)[2ln(3)+3ln(2)]x=2ln(2)−3ln(3) Now you solve for x.