y=xa^(x)=>a^(x-1)= ? A) (y)/(x-1) B) (1)/(x(a-1)) C) (y-1)/(x-1) D) (y)/(ax) E) (y-x)/(x)

Isolate a^x: We are given the equation $ y = x \cdot a^x $. We want to express $ a^{(x-1)} $ in terms of $ y $ and $ x $. To do this, we can start by isolating $ a^x $ on one side of the equation. $ a^x = \frac{y}{x} $Express a^(x-1): Next, we need to express $ a^{(x-1)} $. We can use the property of exponents that states $ a^{(x-1)} = \frac{a^x}{a} $. So, we substitute $ a^x $ with $ \frac{y}{x} $ from the previous step. $ a^{(x-1)} = \frac{\frac{y}{x}}{a} $Simplify the Expression: Now, we simplify the expression by dividing $ \frac{y}{x} $ by $ a $. $ a^{(x-1)} = \frac{y}{ax} $

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y=xa^(x)=>a^(x-1)= ?
A)
(y)/(x-1)
B)
(1)/(x(a-1))
C)
(y-1)/(x-1)
D)
(y)/(ax)
E)
(y-x)/(x)

$37$ . $y=x a^{x} \Rightarrow a^{x-1}=$ ? $\newline$ A) $\frac{y}{x-1}$ $\newline$ B) $\frac{1}{x(a-1)}$ $\newline$ C) $\frac{y-1}{x-1}$ $\newline$ D) $\frac{y}{a x}$ $\newline$ E) $\frac{y-x}{x}$

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Math Problems

Algebra 2

Composition of linear and quadratic functions: find an equation

Full solution

37

y=x a^{x} \Rightarrow a^{x-1}=

\newline

\frac{y}{x-1}

\newline

\frac{1}{x(a-1)}

\newline

\frac{y-1}{x-1}

\newline

\frac{y}{a x}

\newline

\frac{y-x}{x}

Isolate a^x: We are given the equation $y = x \cdot a^x$ . We want to express $a^{(x-1)}$ in terms of $y$ and $x$ . To do this, we can start by isolating $a^x$ on one side of the equation. $\newline$ $a^x = \frac{y}{x}$
Express a^(x $-1$ ): Next, we need to express $a^{(x-1)}$ . We can use the property of exponents that states $a^{(x-1)} = \frac{a^x}{a}$ . $\newline$ So, we substitute $a^x$ with $\frac{y}{x}$ from the previous step. $\newline$ $a^{(x-1)} = \frac{\frac{y}{x}}{a}$
Simplify the Expression: Now, we simplify the expression by dividing $\frac{y}{x}$ by $a$ . $\newline$ $a^{(x-1)} = \frac{y}{ax}$