Solve 9^(x - 1/2) - 4/3^(1 - x) = -1 for x.

Step-by-step solution

1. 1

   Rewrite everything with the same base. Since $9=3^{2}$,

   $9^{\,x-\frac12}=\left(3^{2}\right)^{x-\frac12}=3^{\,2x-1}.$

   Multiplying the exponents is what clears the awkward $\tfrac12$: $2\left(x-\tfrac12\right)=2x-1$ is a whole-number shift.

2. 2

   Move the fraction upstairs with a negative exponent. Because $\frac{1}{3^{\,1-x}}=3^{\,x-1}$,

   $\frac{4}{3^{\,1-x}}=4\cdot3^{\,x-1},$

   and the equation becomes

   $3^{\,2x-1}-4\cdot3^{\,x-1}=-1.$

   Both exponents now differ by a factor of two in the $x$ part, which is the signal that a substitution will linearise the problem.

3. 3

   Substitute $t=3^{\,x-1}$. Then $t>0$ automatically, and

   $3^{\,2x-1}=3^{\,2(x-1)+1}=3\left(3^{\,x-1}\right)^{2}=3t^{2}.$

   Getting the extra factor of $3$ right is the crux: $2x-1$ is $2(x-1)+1$, not $2(x-1)$.

4. 4

   Solve the quadratic. The equation becomes $3t^{2}-4t=-1$, that is

   $3t^{2}-4t+1=0\;\Longrightarrow\;(3t-1)(t-1)=0\;\Longrightarrow\;t=\tfrac13\ \text{or}\ t=1.$

   Both roots are positive, so neither has to be rejected — worth checking, since a negative or zero $t$ could never equal a power of $3$.

5. 5

   Undo the substitution and verify. From $3^{\,x-1}=\tfrac13=3^{-1}$ we get $x-1=-1$, so $x=0$; from $3^{\,x-1}=1=3^{0}$ we get $x=1$. Checking in the original equation: at $x=0$, $9^{-1/2}-\tfrac{4}{3}=\tfrac13-\tfrac43=-1$ ✓; at $x=1$, $9^{1/2}-\tfrac{4}{3^{0}}=3-4=-1$ ✓.