Algebra and Applications Test 2

Concept:For real $x$, the function $f(x)=7^{x-1}+11^{x-1}$ is strictly increasing, so the equation $f(x)=170$ can have at most one real solution.Explanation:Let $f(x)=7^{x-1}+11^{x-1}$.Since $7>1$ and $11>1$, both terms increase as $x$ increases.Hence $f(x)$ is strictly increasing.Check the value $x=3$:$$7^{3-1}+11^{3-1}=7^2+11^2=49+121=170$$So $x=3$ satisfies the equation.Because $f(x)$ is strictly increasing, no other real value of $x$ can give $170$.Answer:The equation has exactly one solution, namely $x=3$.