1. **Problem statement:** We want to find when the graph of $y=\log_a(x)$ intersects the graph of $y=a^x$ at exactly one point, where $a>1$. 2. **Recall definitions:** - $y=\log_a(x)$ is the logarithm base $a$ of $x$, defined for $x>0$. - $y=a^x$ is the exponential function with base $a$. 3. **Set the functions equal to find intersections:** $$\log_a(x) = a^x$$ 4. **Rewrite logarithm in terms of natural log:** $$\log_a(x) = \frac{\ln(x)}{\ln(a)}$$ So the equation becomes: $$\frac{\ln(x)}{\ln(a)} = a^x$$ 5. **Analyze the equation:** - The left side is defined for $x>0$. - The right side is always positive. 6. **Check for the number of solutions:** - At $x=1$, $\log_a(1)=0$ and $a^1=a>1$, so $\log_a(1)1$, there is a unique $x>0$ satisfying these equations. - At this $x$, $y=\log_a(x)$ and $y=a^x$ intersect exactly once (tangentially). **Final answer:** The graphs of $y=\log_a(x)$ and $y=a^x$ intersect at exactly one point when there exists $x>0$ such that $$x \ln(x) = \frac{1}{\ln(a)}$$ and $$a^x = \frac{1}{x (\ln(a))^2}$$ This corresponds to the unique tangential intersection point for $a>1$.