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Math

Math Snap

PROBLEM

(a) 32×3x=813^{-2} \times 3^{x}=81
Find the value of xx.
32×3x=3432×3x3^{-2} \times 3^{x}=3^{4} \quad 3^{-2} \times 3^{x} (x)=4(x)=4
422\frac{4}{2} 2
(b) x13=32x2x^{-\frac{1}{3}}=32 x^{-2}
32+343=3\begin{array}{c} 3^{-2}+3^{4}-3 \\ =3 \end{array} Find the value of xx.

STEP 1

What is this asking?
We need to find the mystery value xx 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 xx in the first equation.
2. Solve for xx in the second equation.

STEP 3

We're given 323x=813^{-2} \cdot 3^{x} = 81.
Since 8181 is 343^4, we can rewrite the equation as 323x=343^{-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, 323x3^{-2} \cdot 3^{x} becomes 32+x3^{-2 + x}.
Our equation is now 32+x=343^{-2 + x} = 3^4.

STEP 5

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

STEP 6

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

STEP 7

We're given x13=32x2x^{-\frac{1}{3}} = 32x^{-2}.
Let's rewrite this to make it easier to work with.
Remember that a negative exponent means "reciprocal," so x2x^{-2} is the same as 1x2\frac{1}{x^2}.
Our equation becomes x13=32x2x^{-\frac{1}{3}} = \frac{32}{x^2}.

STEP 8

To get rid of the fraction, we multiply both sides by x2x^2: x2x13=32x2x2x^2 \cdot x^{-\frac{1}{3}} = \frac{32}{x^2} \cdot x^2.
On the right side, the x2x^2 terms divide to one, leaving us with 3232.

STEP 9

Remember, when multiplying numbers with the same base, we add the exponents.
So, x2x13x^2 \cdot x^{-\frac{1}{3}} becomes x2+(13)x^{2 + (-\frac{1}{3})}, which simplifies to x6313=x53x^{\frac{6}{3} - \frac{1}{3}} = x^{\frac{5}{3}}.
Our equation is now x53=32x^{\frac{5}{3}} = 32.

STEP 10

We know that 3232 is 252^5.
We want to express 3232 as a number raised to the power of 53\frac{5}{3}.
We can write 252^5 as (2553)53(2^{\frac{5}{5} \cdot 3})^{\frac{5}{3}}, which simplifies to (23)53(2^3)^{\frac{5}{3}}.

STEP 11

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

SOLUTION

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

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