## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Re-write as: $49^x=(7^{2})^{x}=(7^{x})^{2}$ We wish to substitute $a=7^{x}$, so the equation becomes: $2a^{2}+11a+5=0$ This gives a quadratic equation, whose factors are: $(2a+1)(a+5)=0$ Use the zero factor property to obtain: $2u+1 =0 \implies a=-\dfrac{1}{2}$ and $a+5 =0 \implies a=-5$ Further, we will plug $a=7^x$ into the above value of $a$ to get the value of $x$ as follows: $7^x=\dfrac{-1}{2}$ and $7^x =-5$. Since, $7^x \gt 0$, the solutions of $7^x=\dfrac{-1}{2}$ and $7^x =-5$ cannot be accepted. Thus, there are no real solutions.