Certainly! Let's go through each expression step-by-step to rewrite them in terms of [tex]\( y = 3^x \)[/tex].
### (a) [tex]\( 3^{3x} \)[/tex]
To express [tex]\( 3^{3x} \)[/tex] in terms of [tex]\( y \)[/tex], we use the properties of exponents. Specifically, we know that:
[tex]\[ 3^{3x} = (3^x)^3 \][/tex]
Since [tex]\( y = 3^x \)[/tex], we substitute [tex]\( y \)[/tex] for [tex]\( 3^x \)[/tex]:
[tex]\[ 3^{3x} = (3^x)^3 = y^3 \][/tex]
Thus:
[tex]\[ 3^{3x} = y^3 \][/tex]
### (b) [tex]\( \frac{1}{3^{x-2}} \)[/tex]
To express [tex]\( \frac{1}{3^{x-2}} \)[/tex] in terms of [tex]\( y \)[/tex], we can simplify this using properties of exponents. First, recall that [tex]\(\frac{1}{a^b} = a^{-b}\)[/tex]:
[tex]\[ \frac{1}{3^{x-2}} = 3^{-(x-2)} \][/tex]
Next, simplify the exponent:
[tex]\[ 3^{-(x-2)} = 3^{-x+2} \][/tex]
This can be rewritten as:
[tex]\[ 3^{-x+2} = 3^{-x} \cdot 3^2 \][/tex]
Using [tex]\( y = 3^x \)[/tex] implies [tex]\( 3^{-x} = \frac{1}{y} \)[/tex]:
[tex]\[ 3^{-x+2} = \frac{1}{y} \cdot 9 = \frac{9}{y} \][/tex]
Thus:
[tex]\[ \frac{1}{3^{x-2}} = \frac{9}{y} \][/tex]
### (c) [tex]\( \frac{81}{9^{2-3x}} \)[/tex]
To express [tex]\( \frac{81}{9^{2-3x}} \)[/tex] in terms of [tex]\( y \)[/tex], we will start by rewriting 81 and 9 in terms of base 3:
[tex]\[ 81 = 3^4 \][/tex]
[tex]\[ 9 = 3^2 \][/tex]
Thus:
[tex]\[ 9^{2-3x} = (3^2)^{2-3x} = 3^{2(2-3x)} = 3^{4-6x} \][/tex]
Now, rewrite the entire expression:
[tex]\[ \frac{81}{9^{2-3x}} = \frac{3^4}{3^{4-6x}} \][/tex]
Using the properties of exponents, specifically [tex]\( \frac{a^m}{a^n} = a^{m-n} \)[/tex], we get:
[tex]\[ \frac{3^4}{3^{4-6x}} = 3^{4-(4-6x)} = 3^{4-4+6x} = 3^{6x} \][/tex]
Since [tex]\( y = 3^x \)[/tex], we have:
[tex]\[ 3^{6x} = (3^x)^6 = y^6 \][/tex]
Thus:
[tex]\[ \frac{81}{9^{2-3x}} = y^6 \][/tex]
In conclusion, the expressions in their simplest forms in terms of [tex]\( y \)[/tex] are:
(a) [tex]\( 3^{3x} = y^3 \)[/tex]
(b) [tex]\( \frac{1}{3^{x-2}} = \frac{9}{y} \)[/tex]
(c) [tex]\( \frac{81}{9^{2-3x}} = y^6 \)[/tex]
