Solve for x and y y4logx5 y32x2 Solve for x and y y4logx5 y3

Solution

given y+3 = 2 X^2 ----- 1 and y-4 = log(x+5) -------2

from equation 1 we have y = 2 X^2 - 3 substitute this in 2

2X^2 - 3 - 4 = log ( x+5)

2X^2 - 7 = log(x+5)

2X^2 - log(x+2) - 7 = 0

for log(a+b) we have loga + log (1+b)/a substituting this in above equation

2X^2 - (log x + log ( 1+ 2 / x ) - 7 =0

2 X^2 - log x - log(3/x) - 7 =0

2 X^2 - log x - (log 3 - log x) - 7 = 0

2 X^2 -logx - log 3 + log x -7 = 0

2X^2 - log 3 -7 =0

the value of log 3 is 0.477 substituting

2 X^2 - 0.477 - 7 = 0

2X^2 = 7.477

X^2 = 3.7385

X= root ( 3.7385)

X = 1.933

substitute this in second equation

y - 4 = log ( 1.933+5)

y = log(6.933) + 4

Y = 0.8409 + 4

Y = 4.8409

Hence X = 1.933 and   Y = 4.8409