Solved on Dec 12, 2023

Solve the equation: $\left(\frac{1}{16}\right)^{-x}=\left(\frac{1}{32}\right)^{-x-1}$

STEP 1

Assumptions
1. We are given the equation $\left(\frac{1}{16}\right)^{-x}=\left(\frac{1}{32}\right)^{-x-1}$ .
2. We need to solve for the variable $x$ .

STEP 2

First, we will rewrite the bases of the exponents as powers of 2, since both 16 and 32 are powers of 2.
$16 = 2^4$ $32 = 2^5$

STEP 3

Now, rewrite the equation using the new bases.
$\left(\frac{1}{2^4}\right)^{-x}=\left(\frac{1}{2^5}\right)^{-x-1}$

STEP 4

Next, we apply the negative exponent rule, which states that $a^{-n} = \frac{1}{a^n}$ , to both sides of the equation.
$\left(2^4\right)^{x}=\left(2^5\right)^{x+1}$

STEP 5

Now, apply the power of a power rule, which states that $(a^m)^n = a^{mn}$ , to both sides of the equation.
$2^{4x}=2^{5(x+1)}$

STEP 6

Simplify the right side of the equation by distributing the exponent.
$2^{4x}=2^{5x+5}$

STEP 7

Since the bases are the same on both sides of the equation, we can set the exponents equal to each other.
$4x = 5x + 5$

STEP 8

Now, we will solve for $x$ by isolating it on one side of the equation. First, subtract $5x$ from both sides.
$4x - 5x = 5x + 5 - 5x$

STEP 9

Simplify the equation by combining like terms.
$-x = 5$

STEP 10

Finally, multiply both sides of the equation by $-1$ to solve for $x$ .
$x = -5$
The solution to the equation $\left(\frac{1}{16}\right)^{-x}=\left(\frac{1}{32}\right)^{-x-1}$ is $x = -5$ .

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Solve the equation: $\left(\frac{1}{16}\right)^{-x}=\left(\frac{1}{32}\right)^{-x-1}$

STEP 1

Assumptions
1. We are given the equation $\left(\frac{1}{16}\right)^{-x}=\left(\frac{1}{32}\right)^{-x-1}$ .
2. We need to solve for the variable $x$ .

STEP 2

First, we will rewrite the bases of the exponents as powers of 2, since both 16 and 32 are powers of 2.
$16 = 2^4$ $32 = 2^5$

STEP 3

Now, rewrite the equation using the new bases.
$\left(\frac{1}{2^4}\right)^{-x}=\left(\frac{1}{2^5}\right)^{-x-1}$

STEP 4

Next, we apply the negative exponent rule, which states that $a^{-n} = \frac{1}{a^n}$ , to both sides of the equation.
$\left(2^4\right)^{x}=\left(2^5\right)^{x+1}$

STEP 5

Now, apply the power of a power rule, which states that $(a^m)^n = a^{mn}$ , to both sides of the equation.
$2^{4x}=2^{5(x+1)}$

STEP 6

Simplify the right side of the equation by distributing the exponent.
$2^{4x}=2^{5x+5}$

STEP 7

Since the bases are the same on both sides of the equation, we can set the exponents equal to each other.
$4x = 5x + 5$

STEP 8

Now, we will solve for $x$ by isolating it on one side of the equation. First, subtract $5x$ from both sides.
$4x - 5x = 5x + 5 - 5x$

STEP 9

Simplify the equation by combining like terms.
$-x = 5$

STEP 10

Finally, multiply both sides of the equation by $-1$ to solve for $x$ .
$x = -5$
The solution to the equation $\left(\frac{1}{16}\right)^{-x}=\left(\frac{1}{32}\right)^{-x-1}$ is $x = -5$ .

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Solve the equation: (116)−x=(132)−x−1\left(\frac{1}{16}\right)^{-x}=\left(\frac{1}{32}\right)^{-x-1}(161​)−x=(321​)−x−1

STEP 1

Assumptions1. We are given the equation (116)−x=(132)−x−1\left(\frac{1}{16}\right)^{-x}=\left(\frac{1}{32}\right)^{-x-1}(161​)−x=(321​)−x−1.2. We need to solve for the variable xxx.

STEP 2

First, we will rewrite the bases of the exponents as powers of 2, since both 16 and 32 are powers of 2.16=2416 = 2^416=24 32=2532 = 2^532=25

STEP 3

Now, rewrite the equation using the new bases.(124)−x=(125)−x−1\left(\frac{1}{2^4}\right)^{-x}=\left(\frac{1}{2^5}\right)^{-x-1}(241​)−x=(251​)−x−1

STEP 4

Next, we apply the negative exponent rule, which states that a−n=1ana^{-n} = \frac{1}{a^n}a−n=an1​, to both sides of the equation.(24)x=(25)x+1\left(2^4\right)^{x}=\left(2^5\right)^{x+1}(24)x=(25)x+1

STEP 5

Now, apply the power of a power rule, which states that (am)n=amn(a^m)^n = a^{mn}(am)n=amn, to both sides of the equation.24x=25(x+1)2^{4x}=2^{5(x+1)}24x=25(x+1)

STEP 6

Simplify the right side of the equation by distributing the exponent.24x=25x+52^{4x}=2^{5x+5}24x=25x+5

STEP 7

Since the bases are the same on both sides of the equation, we can set the exponents equal to each other.4x=5x+54x = 5x + 54x=5x+5

STEP 8

Now, we will solve for xxx by isolating it on one side of the equation. First, subtract 5x5x5x from both sides.4x−5x=5x+5−5x4x - 5x = 5x + 5 - 5x4x−5x=5x+5−5x

STEP 9

Simplify the equation by combining like terms.−x=5-x = 5−x=5

STEP 10

Finally, multiply both sides of the equation by −1-1−1 to solve for xxx.x=−5x = -5x=−5The solution to the equation (116)−x=(132)−x−1\left(\frac{1}{16}\right)^{-x}=\left(\frac{1}{32}\right)^{-x-1}(161​)−x=(321​)−x−1 is x=−5x = -5x=−5.

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Solve the equation: $\left(\frac{1}{16}\right)^{-x}=\left(\frac{1}{32}\right)^{-x-1}$

Assumptions
1. We are given the equation $\left(\frac{1}{16}\right)^{-x}=\left(\frac{1}{32}\right)^{-x-1}$ .
2. We need to solve for the variable $x$ .

First, we will rewrite the bases of the exponents as powers of 2, since both 16 and 32 are powers of 2.
$16 = 2^4$ $32 = 2^5$

Now, rewrite the equation using the new bases.
$\left(\frac{1}{2^4}\right)^{-x}=\left(\frac{1}{2^5}\right)^{-x-1}$

Next, we apply the negative exponent rule, which states that $a^{-n} = \frac{1}{a^n}$ , to both sides of the equation.
$\left(2^4\right)^{x}=\left(2^5\right)^{x+1}$

Now, apply the power of a power rule, which states that $(a^m)^n = a^{mn}$ , to both sides of the equation.
$2^{4x}=2^{5(x+1)}$

Simplify the right side of the equation by distributing the exponent.
$2^{4x}=2^{5x+5}$

Since the bases are the same on both sides of the equation, we can set the exponents equal to each other.
$4x = 5x + 5$

Now, we will solve for $x$ by isolating it on one side of the equation. First, subtract $5x$ from both sides.
$4x - 5x = 5x + 5 - 5x$

Simplify the equation by combining like terms.
$-x = 5$

Finally, multiply both sides of the equation by $-1$ to solve for $x$ .
$x = -5$
The solution to the equation $\left(\frac{1}{16}\right)^{-x}=\left(\frac{1}{32}\right)^{-x-1}$ is $x = -5$ .