SOLUTION: solve for x and y ; if x>0 and y>0 : log xy=log x/y + 2 log 2= 2
Algebra.Com
Question 496065: solve for x and y ; if x>0 and y>0 : log xy=log x/y + 2 log 2= 2
Answer by chessace(471) (Show Source): You can put this solution on YOUR website!
log x + log y = 2
log x - log y + 2 log 2 = 2
2 log y - 2 log 2 = 0
y = 2
log 2x = 2
2x = 100
x = 50
RELATED QUESTIONS
Solve this simultaneous equation for x and y.
log(base2)X + log(base2)Y =2... (answered by josgarithmetic)
Please help me solve this equation: if log x + log y =2, find... (answered by lwsshak3)
please help me solve this equation: if log x + log y = 2, find... (answered by lwsshak3)
Please help me solve this problem:
If xy = 64 and {{{ log (x, y) + log (y, x) = 5/2... (answered by ikleyn)
x² + y² = 23xy and x,y > 0 prove that log({{{(x + y)/5}}}] = {{{1/2}}}[log(x) +... (answered by Edwin McCravy)
If {{{ log((x + y)/5)}}} = {{{ log(sqrt(x))+ log(sqrt(y)) }}} where x and y are greater... (answered by solver91311,lwsshak3)
if {{{ log( b, x) = 6 }}} and {{{ log( b, y) = 2 }}} then solve {{{ log( b, (x^2)/y) }}}
(answered by scott8148,lwsshak3,Alan3354)
Solve the simultaneous equations.
log base 2 (xy) = 7
log base 2 (x^2/y) =... (answered by josgarithmetic)
Please help me solve for x and y:
(log(3,x))(log(x,2x))(log(2x,y)) = log(x,y^2)... (answered by Theo)