Math / Algebra

Question(a) $3^{-2} \times 3^{x}=81$
Find the value of $x$ . $3^{-2} \times 3^{x}=3^{4} \quad 3^{-2} \times 3^{x}$ $(x)=4$ $\frac{4}{2} 2$ (b) $x^{-\frac{1}{3}}=32 x^{-2}$ $\begin{array}{c} 3^{-2}+3^{4}-3 \\ =3 \end{array}$
Find the value of $x$ .

Studdy Solution

STEP 1

What is this asking? We need to find the mystery value $x$ that makes two equations true, one where powers of 3 are multiplied, and another involving fractional powers and reciprocals. Watch out! Remember your exponent rules!
Don't mix up adding and multiplying exponents.
Also, be careful with those negative and fractional exponents; they can be tricky!

STEP 2

1. Solve for $x$ in the first equation.
2. Solve for $x$ in the second equation.

STEP 3

We're given $3^{-2} \cdot 3^{x} = 81$ .
Since $81$ is $3^4$ , we can rewrite the equation as $3^{-2} \cdot 3^{x} = 3^4$ .
This makes it easier to compare the exponents since they have the same base.

STEP 4

When multiplying numbers with the same base, we add the exponents.
So, $3^{-2} \cdot 3^{x}$ becomes $3^{-2 + x}$ .
Our equation is now $3^{-2 + x} = 3^4$ .

STEP 5

Since the bases are the same, the exponents must be equal.
This gives us $-2 + x = 4$ .

STEP 6

To isolate $x$ , we add 2 to both sides of the equation: $-2 + x + 2 = 4 + 2$ .
This simplifies to $x = \textbf{6}$ .

STEP 7

We're given $x^{-\frac{1}{3}} = 32x^{-2}$ .
Let's rewrite this to make it easier to work with.
Remember that a negative exponent means "reciprocal," so $x^{-2}$ is the same as $\frac{1}{x^2}$ .
Our equation becomes $x^{-\frac{1}{3}} = \frac{32}{x^2}$ .

STEP 8

To get rid of the fraction, we multiply both sides by $x^2$ : $x^2 \cdot x^{-\frac{1}{3}} = \frac{32}{x^2} \cdot x^2$ .
On the right side, the $x^2$ terms divide to one, leaving us with $32$ .

STEP 9

Remember, when multiplying numbers with the same base, we add the exponents.
So, $x^2 \cdot x^{-\frac{1}{3}}$ becomes $x^{2 + (-\frac{1}{3})}$ , which simplifies to $x^{\frac{6}{3} - \frac{1}{3}} = x^{\frac{5}{3}}$ .
Our equation is now $x^{\frac{5}{3}} = 32$ .

STEP 10

We know that $32$ is $2^5$ .
We want to express $32$ as a number raised to the power of $\frac{5}{3}$ .
We can write $2^5$ as $(2^{\frac{5}{5} \cdot 3})^{\frac{5}{3}}$ , which simplifies to $(2^3)^{\frac{5}{3}}$ .

STEP 11

Now we have $x^{\frac{5}{3}} = (2^3)^{\frac{5}{3}}$ .
Since the exponents are the same, we can equate the bases: $x = 2^3$ .
Therefore, $x = \textbf{8}$ .

STEP 12

For the first equation, $x = \textbf{6}$ .
For the second equation, $x = \textbf{8}$ .

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Question(a) $3^{-2} \times 3^{x}=81$
Find the value of $x$ . $3^{-2} \times 3^{x}=3^{4} \quad 3^{-2} \times 3^{x}$ $(x)=4$ $\frac{4}{2} 2$ (b) $x^{-\frac{1}{3}}=32 x^{-2}$ $\begin{array}{c} 3^{-2}+3^{4}-3 \\ =3 \end{array}$
Find the value of $x$ .

Studdy Solution

STEP 1

STEP 2

1. Solve for $x$ in the first equation.
2. Solve for $x$ in the second equation.

STEP 3

We're given $3^{-2} \cdot 3^{x} = 81$ .
Since $81$ is $3^4$ , we can rewrite the equation as $3^{-2} \cdot 3^{x} = 3^4$ .
This makes it easier to compare the exponents since they have the same base.

STEP 4

When multiplying numbers with the same base, we add the exponents.
So, $3^{-2} \cdot 3^{x}$ becomes $3^{-2 + x}$ .
Our equation is now $3^{-2 + x} = 3^4$ .

STEP 5

Since the bases are the same, the exponents must be equal.
This gives us $-2 + x = 4$ .

STEP 6

To isolate $x$ , we add 2 to both sides of the equation: $-2 + x + 2 = 4 + 2$ .
This simplifies to $x = \textbf{6}$ .

STEP 7

We're given $x^{-\frac{1}{3}} = 32x^{-2}$ .
Let's rewrite this to make it easier to work with.
Remember that a negative exponent means "reciprocal," so $x^{-2}$ is the same as $\frac{1}{x^2}$ .
Our equation becomes $x^{-\frac{1}{3}} = \frac{32}{x^2}$ .

STEP 8

To get rid of the fraction, we multiply both sides by $x^2$ : $x^2 \cdot x^{-\frac{1}{3}} = \frac{32}{x^2} \cdot x^2$ .
On the right side, the $x^2$ terms divide to one, leaving us with $32$ .

STEP 9

Remember, when multiplying numbers with the same base, we add the exponents.
So, $x^2 \cdot x^{-\frac{1}{3}}$ becomes $x^{2 + (-\frac{1}{3})}$ , which simplifies to $x^{\frac{6}{3} - \frac{1}{3}} = x^{\frac{5}{3}}$ .
Our equation is now $x^{\frac{5}{3}} = 32$ .

STEP 10

We know that $32$ is $2^5$ .
We want to express $32$ as a number raised to the power of $\frac{5}{3}$ .
We can write $2^5$ as $(2^{\frac{5}{5} \cdot 3})^{\frac{5}{3}}$ , which simplifies to $(2^3)^{\frac{5}{3}}$ .

STEP 11

Now we have $x^{\frac{5}{3}} = (2^3)^{\frac{5}{3}}$ .
Since the exponents are the same, we can equate the bases: $x = 2^3$ .
Therefore, $x = \textbf{8}$ .

STEP 12

For the first equation, $x = \textbf{6}$ .
For the second equation, $x = \textbf{8}$ .

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Studdy Solution

STEP 1

STEP 2

1. Solve for xxx in the first equation.2. Solve for xxx in the second equation.

STEP 3

We're given 3−2⋅3x=813^{-2} \cdot 3^{x} = 813−2⋅3x=81.Since 818181 is 343^434, we can rewrite the equation as 3−2⋅3x=343^{-2} \cdot 3^{x} = 3^43−2⋅3x=34.This makes it easier to compare the exponents since they have the same base.

STEP 4

When multiplying numbers with the same base, we *add* the exponents.So, 3−2⋅3x3^{-2} \cdot 3^{x}3−2⋅3x becomes 3−2+x3^{-2 + x}3−2+x.Our equation is now 3−2+x=343^{-2 + x} = 3^43−2+x=34.

STEP 5

Since the bases are the same, the exponents must be equal.This gives us −2+x=4-2 + x = 4−2+x=4.

STEP 6

To isolate xxx, we add **2** to both sides of the equation: −2+x+2=4+2-2 + x + 2 = 4 + 2−2+x+2=4+2.This simplifies to x=6x = \textbf{6}x=6.

STEP 7

STEP 8

To get rid of the fraction, we multiply both sides by x2x^2x2: x2⋅x−13=32x2⋅x2x^2 \cdot x^{-\frac{1}{3}} = \frac{32}{x^2} \cdot x^2x2⋅x−31​=x232​⋅x2.On the right side, the x2x^2x2 terms divide to one, leaving us with 323232.

STEP 9

STEP 10

We know that 323232 is 252^525.We want to express 323232 as a number raised to the power of 53\frac{5}{3}35​.We can write 252^525 as (255⋅3)53(2^{\frac{5}{5} \cdot 3})^{\frac{5}{3}}(255​⋅3)35​, which simplifies to (23)53(2^3)^{\frac{5}{3}}(23)35​.

STEP 11

Now we have x53=(23)53x^{\frac{5}{3}} = (2^3)^{\frac{5}{3}}x35​=(23)35​.Since the exponents are the same, we can equate the bases: x=23x = 2^3x=23.Therefore, x=8x = \textbf{8}x=8.

STEP 12

For the first equation, x=6x = \textbf{6}x=6.For the second equation, x=8x = \textbf{8}x=8.

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1. Solve for $x$ in the first equation.
2. Solve for $x$ in the second equation.

We're given $3^{-2} \cdot 3^{x} = 81$ .
Since $81$ is $3^4$ , we can rewrite the equation as $3^{-2} \cdot 3^{x} = 3^4$ .
This makes it easier to compare the exponents since they have the same base.

When multiplying numbers with the same base, we add the exponents.
So, $3^{-2} \cdot 3^{x}$ becomes $3^{-2 + x}$ .
Our equation is now $3^{-2 + x} = 3^4$ .

Since the bases are the same, the exponents must be equal.
This gives us $-2 + x = 4$ .

To isolate $x$ , we add 2 to both sides of the equation: $-2 + x + 2 = 4 + 2$ .
This simplifies to $x = \textbf{6}$ .

To get rid of the fraction, we multiply both sides by $x^2$ : $x^2 \cdot x^{-\frac{1}{3}} = \frac{32}{x^2} \cdot x^2$ .
On the right side, the $x^2$ terms divide to one, leaving us with $32$ .

We know that $32$ is $2^5$ .
We want to express $32$ as a number raised to the power of $\frac{5}{3}$ .
We can write $2^5$ as $(2^{\frac{5}{5} \cdot 3})^{\frac{5}{3}}$ , which simplifies to $(2^3)^{\frac{5}{3}}$ .

Now we have $x^{\frac{5}{3}} = (2^3)^{\frac{5}{3}}$ .
Since the exponents are the same, we can equate the bases: $x = 2^3$ .
Therefore, $x = \textbf{8}$ .

For the first equation, $x = \textbf{6}$ .
For the second equation, $x = \textbf{8}$ .