Question

16. Solve for x in each of the following (hint: logs or factoring): A. 32-x-4 = 16-2x+1 B. x4 +7x3 +12x2 = 0

Answer 1

SOLUTION:

Case: Exponential equations

Method:

a)

$\begin{gathered} 32^{-x-4}=16^{-2x+1} \\ 2^{5(-x-4)}=2^{4(-2x+1)} \\ 5(-x-4)=4(-2x+1) \\ -5x-20=-8x+4 \\ -5x+8x=4+20 \\ 3x=24 \\ x=\frac{24}{3} \\ x=8 \end{gathered}$

b)

$\begin{gathered} x^4+7x^3+12x^2=0 \\ Factorizing \\ x^2(x^2+7x+12)=0 \\ x^2(x^2+4x+3x+12)=0 \\ x^2[x(x+4)+3(x+4)]=0 \\ x^2(x+4)(x+3)=0 \\ x=0,\text{ }x=-4\text{ OR }x=-3 \end{gathered}$

Final answer:

a) x= 8

b) x= 0 or -4 or -3