solve the equation |4^(y+1) - 1| = |4^y -3|

Question

need the step-by-step method asap. your help is very much needed and appreciated. thanks! only God can repay your kind deed.

Answer 1

If lAl = lBl, then A = B or A = -B

l 4^y+1 - 1l = l4^y-3l

 

Case 1:  4^y+1-1 = 4^y-3

 

            4(4^y) - 1 = 4^y - 3

 

            4(4^y) - 4^y = -2

 

            4^y(4-1) = -2

 

              3(4^y) = -2

 

            This equation has no solution because the left side of the                   equation is positive for all values of y.

 

Case 2:  4^y+1 - 1 = -(4^y - 3)

 

            4(4^y) - 1 = -4^y + 3

 

            4(4^y) + 4^y = 4

 

             4^y(4+1) = 4

 

             5(4^y) = 4

 

                4^y = 4/5

 

             log(4^y) = 4/5

 

             y(log4) = log(4/5)

 

                    y = log(4/5) / log4 ≈ -0.160964047

 

                

Answer 2

I would remove the absolute value symbols.  This is because the absolute value of anything regardless of signs, is always positive.

 

4^{(y + 1)} - 1 = 4^y - 3

 

 

Now we can solve for y.

 

On the left side of equation, apply the rules of exponents.

 

4^y(4) - 1 = 4^y - 3

 

 

Move all the 4^y terms to the left side of equation.  Move all the constant terms to the right side.

 

4^y(4) - 4^y = -2

 

 

Factor out a 4^y.

 

4^y(4 - 1) = -2

 

4^y = -2/3

 

 

Log both sides of the equation.  Bring any exponents as the coefficient of the log.

 

ylog(4) = log(-2/3)

 

y = log(-2/3) / log(4)

 

 

Because you cannot have a negative argument for log, there is no solution.