2

Find the possible values of $x$ if $2^{2x+1} = 3(2^x) -1$

I know that $x=0$ and $x=-1$ are possible values of $x$ by looking at the equation. I need help understanding how to use logarithms to solve questions of this type. Here is what I'm doing, where am I going wrong?:

$$2^{2x+1} = 3(2^x) -1$$ Can be written as $$ 3(2^x) - 2^{2x+1} =1$$ Taking logarithms of each side (and here is where I think I go wrong): $$[x \ln(2) + \ln(3)] - [2x \ln(2) + \ln(2)] = \ln(1)$$ $$[x \ln(2) + \ln(3)] - [2x \ln(2) + \ln(2)] = 0$$

$$-x \ln(2) + \ln(3) - \ln(2) = 0$$

mikoyan
  • 1,135

1 Answers1

6

$$2^{2x+1} = 3(2^x) -1$$ $$2\cdot2^{2x}-3\cdot2^x+1=0$$ $2^x=t>0$ $$2t^2-3t+1=0$$ $t=1$ or $t=\frac12$

$2^x=2^{-1}$ or $2^x=1$

$x=-1$ or $x=0$

Roman83
  • 17,884
  • 3
  • 26
  • 70