Solve for x log of 4x^2+5x = log of 36x+90

Problem

(log_)(4*x2+5*x)=(log_)(36*x+90)

Solution

1. Apply the property of equality for logarithms, which states that if (log_)(A)=(log_)(B) then A=B

4*x2+5*x=36*x+90

2. Rearrange the equation into standard quadratic form a*x2+b*x+c=0 by subtracting 36*x and 90 from both sides.

4*x2−31*x−90=0

3. Factor the quadratic equation by finding two numbers that multiply to 4⋅(−90)=−360 and add to −31 These numbers are −40 and 9

4*x2−40*x+9*x−90=0

4*x*(x−10)+9*(x−10)=0

(4*x+9)*(x−10)=0

4. Solve for x by setting each factor equal to zero.

4*x+9=0⇒x=−9/4

x−10=0⇒x=10

5. Check for extraneous solutions by ensuring the arguments of the original logarithms are positive. For x=10 both 4*(10)2+5*(10) and 36(10) + 90a*r*e*p*o*s(i)*t*i*v*e.F*o*r = -2.25,(-2.25)^2 + 5(-2.25) = 20.25 - 11.25 = 9$, which is positive. Both solutions are valid.

Final Answer

(log_)(4*x2+5*x)=(log_)(36*x+90)⇒x=−9/4,10

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