The functions $f(x)=8\left(\frac{2}{5}\right)^{x}$ and $g(x)=8(b)^{x}$ are graphed in the $x y$ -plane. For what value of $b$ would the graphs of functions $f$ and $g$ be symmetric with respect to the $y$ -axis?

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Algebra 2

Domain and range of quadratic functions: equations

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Q. The functions

f(x)=8\left(\frac{2}{5}\right)^{x}

and

g(x)=8(b)^{x}

are graphed in the

x y

-plane. For what value of

b

would the graphs of functions

f

and

g

be symmetric with respect to the

y

-axis?

Find $b$ for symmetry: To find the value of $b$ for which the graphs of functions g(x) are symmetric with respect to the y-axis, we need to ensure that $f(x) = g(-x)$ .
Express $g(-x)$ : Let's first express $g(-x)$ using the function $g(x) = 8b^x$ . $\newline$ $g(-x) = 8b^{-x} = \frac{8}{b^x}$
Set the equation: Now we have, $f(x) = 8(\frac{2}{5})^x$ and $g(-x) = 8b^{-x}$ $\newline$ We can set them equal to each other. $\newline$ $8(\frac{2}{5})^x = 8b^{-x}$ $\newline$ Let's simplify this equation, by multiplying by $8$ on both sides of the equation. $\newline$ $(\frac{2}{5})^x = b^{-x}$
Eliminate the exponents: To eliminate the exponents, take logarithm on both sides. $\newline$ $\text{log}((\frac{2}{5})^x) = \text{log}(b^{-x})$ $\newline$ Using the properties of logarithms, Power rule: $\text{log}(a^b) = b \text{log}(a)$ $\newline$ $x\text{log}(\frac{2}{5}) = -x \text{log}(b)$
Solve for $b$ : Now, we divide both sides by $-x$ and $\text{log}(\frac{2}{5})$ to isolate $b$ term. $\newline$ $\frac{x}{-x} = \frac{\text{log}(b)}{\text{log}(\frac{2}{5})}$ $\newline$ Let's start simplifying the equation: $\newline$ $-1 = \frac{\text{log}(b)}{\text{log}(\frac{2}{5})}$ $\newline$ Eliminate the fraction using cross-multiplication. $\newline$ $-\text{log}(\frac{2}{5}) = \text{log}(b)$ $\newline$ Using the properties of logarithms, Negative rule: $-\text{log}(\frac{a}{b}) = \text{log}(\frac{b}{a})$ $\newline$ $\text{log}(\frac{5}{2}) = \text{log}(b)$ $\newline$ Which implies, $b = \frac{5}{2}$ .