Solve the following exponential equations:
a) $$6^x-9 \cdot 6^{-x}+8=0$$
b) $$3^{2(x+1)}-18\cdot 3^x+9=0$$
See development and solution
Development:
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 $$$9t^2-18t+9=0 \Rightarrow t=1; t=1$$$ $$$1=3^x \Rightarrow x=0$$$
Solution:
a) $$x=0$$
b) $$x=0$$