Solving an exponential equation by variable change
Solve the following exponential equations:
a) $6^x-9 \cdot 6^{-x}+8=0$
b) $3^{2(x+1)}-18\cdot 3^x+9=0$
a) $6^x-9 \cdot 6^{-x}+8=0$
We multiply the whole expression by $6^x$ and use the variable change $t=6^x$ $$6^{2x}-9+8\cdot6^x=0 \Rightarrow t^2+8t-9=0 \Rightarrow t=1;t=-9$$ Then, one takes the positive solution, since the negative one would not make sense when applying the logarithm. We have $$x=log_6 1=0$$
b) $3^{2(x+1)}-18\cdot 3^x+9=0$
The variable change $t=3^x$ is introduced and one gets $$9t^2-18t+9=0 \Rightarrow t=1; t=1$$ $$1=3^x \Rightarrow x=0$$
a) $x=0$
b) $x=0$