Question

Deduce x in terms of y for the following.

x^2-529y^4=(log10^3)^2+2(log10^6/2)(log10^45/log10^3+log100^4)y^2

Answer 1

We know that

1) log10 = 1 and
2) logx^y =ylogx.

First simplifying the R.H.S

x^2-529y^4=(log10^3)^2+2(log10^6/2)(log10^45/log10^3+log100^4)y^2

x^2-529y^4=(3log10)^2+2(log10^3) ( 45log10/3log10 +(4 log10^2)y^2

x^2-529y^4=3^2+2(3)(45/3+8)y^2

x^2-529y^4=(3)^2+2(3)(23y^2)

x^2= (3)^2 +2(3)(23y^2) +529y^4

x^2= (3)^2 +2(3)(23y^2) +(23y^2)^2

This is now in the form of (a+b)^2

x^2 = (3+23y^2) ^2

Applying square root on both sides we get.

x =± 3+ 23y^2

x=3+23y²

x=-3-23y²

Therefore, x in terms of y => x = ±3+23y^2

Answer 2

Answer: x = 23 y^2 + 3, or , -23 y^2 -3.

two values. We know Log 10^n = n Log 10 = n., log 10 = 1.

we also know how to solve a quadratic equation with discriminant. .