Sure, let's solve the problem step by step.
Given:
[tex]\[ \log_2 x = a \][/tex]
[tex]\[ \log_5 y = a \][/tex]
We need to express [tex]\( 100^{2a - 1} \)[/tex] in terms of [tex]\( x \)[/tex] and [tex]\( y \)[/tex].
### Step 1: Express [tex]\( a \)[/tex] in exponential form
From [tex]\(\log_2 x = a\)[/tex], we can express [tex]\( x \)[/tex] as:
[tex]\[ x = 2^a \][/tex]
Similarly, from [tex]\(\log_5 y = a\)[/tex], we can express [tex]\( y \)[/tex] as:
[tex]\[ y = 5^a \][/tex]
### Step 2: Rewrite [tex]\( 100 \)[/tex] in base terms
We know:
[tex]\[ 100 = 10^2 = (2 \times 5)^2 = 2^2 \times 5^2 \][/tex]
### Step 3: Rewrite the expression [tex]\( 100^{2a-1} \)[/tex]
[tex]\[ 100^{2a - 1} = \left(2^2 \times 5^2\right)^{2a - 1} \][/tex]
### Step 4: Apply the power rule
Using the rule [tex]\((ab)^n = a^n \times b^n\)[/tex]:
[tex]\[ \left(2^2 \times 5^2\right)^{2a - 1} = (2^2)^{2a - 1} \times (5^2)^{2a - 1} \][/tex]
### Step 5: Simplify powers
[tex]\[ = 2^{2(2a - 1)} \times 5^{2(2a - 1)} \][/tex]
[tex]\[ = 2^{4a - 2} \times 5^{4a - 2} \][/tex]
### Step 6: Substitute back [tex]\( a \)[/tex]:
From [tex]\( x = 2^a \)[/tex] so [tex]\( a = \log_2 x \)[/tex], hence [tex]\( 2^{4a - 2} = (2^a)^{4 - \frac{2}{a}} = x^{4 - \frac{2}{a}} \)[/tex]:
[tex]\[ 2^{4a - 2} = (2^a)^{4 - \frac{2}{a}} = 2^{4a - 2} = x^4 \][/tex]
Similarly for [tex]\(y = 5^a \)[/tex] so [tex]\( a = \log_5 y \)[/tex]:
[tex]\[2^{4a - 2} = (5^a) ^{4 - \frac{2}{a}} = y^{4 - \frac{2}{a}} \][/tex]
### Step 7: Plug these back into the expression
[tex]\[ \therefore 100^{2a - 1} = x^4 \times y^4 = (xy)^4 \][/tex]
Thus, [tex]\( 100^{2a - 1} \)[/tex] in terms of [tex]\( x \)[/tex] and [tex]\( y \)[/tex] is:
[tex]\[ 100^{2a - 1} = (xy)^4 \][/tex]
And that is your final answer.
