Question

match each equation with a graph above
$10^{x}$ a. red (r)
$\log(x)$ b. black (k)
$e^{x}$ c. blue (b)
$\ln(x)$ d. green (g)

Explanation:

• For \(y = 10^{x}\):
• The function \(y = a^{x}\) with \(a>1\) (here \(a = 10\)) is an exponential - growth function.
• The blue curve (B) has a steeper growth rate compared to \(y=e^{x}\) for positive \(x\) values. Since \(10>e\approx2.718\), \(y = 10^{x}\) grows faster than \(y=e^{x}\) for \(x>0\).
• For \(y=\log(x)\):
• The function \(y = \log(x)=\log_{10}(x)\) has a slower growth rate than \(y=\ln(x)\) for \(x > 1\).
• The green curve (G) has a relatively slower - increasing trend for \(x>1\) compared to the black curve.
• For \(y = e^{x}\):
• The function \(y = e^{x}\) is an exponential - growth function.
• The red curve (R) has a growth rate that is less than \(y = 10^{x}\) (since \(e<10\)) but still shows exponential - like growth (passes through \((0,1)\) and is increasing for all \(x\)).
• For \(y=\ln(x)\):
• The function \(y=\ln(x)\) is the natural - logarithm function.
• The black curve (K) has a domain \(x>0\), passes through \((1,0)\), and has a relatively steeper increase near \(x = 1\) compared to \(y=\log(x)\) (because the derivative of \(y=\ln(x)\) is \(y'=\frac{1}{x}\) and the derivative of \(y=\log(x)=\frac{\log_{10}(e)}{x}\approx\frac{0.434}{x}\), and \(1>0.434\)).

Answer:

1. \(10^{x}\): c. blue (B)
2. \(\log(x)\): d. green (G)
3. \(e^{x}\): a. red (R)
4. \(\ln(x)\): b. black (K)