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Sagot :
To prove the equation [tex]\(\frac{1}{z} = \frac{3}{x} + \frac{1}{y}\)[/tex] given the relations [tex]\(2^x = 5\)[/tex] and [tex]\(y = 40z\)[/tex], follow these steps:
1. Expression for [tex]\(z\)[/tex] in terms of [tex]\(y\)[/tex]:
Given [tex]\(y = 40z\)[/tex], we can write:
[tex]\[ z = \frac{y}{40} \][/tex]
2. Substitute [tex]\(z\)[/tex] into the equation to be proven:
Substitute [tex]\(z\)[/tex] as [tex]\(\frac{y}{40}\)[/tex] in the equation:
[tex]\[ \frac{1}{z} = \frac{40}{y} \][/tex]
3. Rewrite the equation:
Now, we need to prove:
[tex]\[ \frac{40}{y} = \frac{3}{x} + \frac{1}{y} \][/tex]
4. Isolate the common denominator:
Multiply both sides of the equation by [tex]\(y\)[/tex] to clear the fraction:
[tex]\[ 40 = 3 \cdot \frac{y}{x} + 1 \][/tex]
Simplify:
[tex]\[ 40 = 3 \cdot \frac{y}{x} + 1 \][/tex]
5. Rearrange the terms:
Move 1 to the left side:
[tex]\[ 40 - 1 = 3 \cdot \frac{y}{x} \][/tex]
Simplify:
[tex]\[ 39 = 3 \cdot \frac{y}{x} \][/tex]
6. Isolate [tex]\(x\)[/tex]:
Rearrange to solve for [tex]\(x\)[/tex]:
[tex]\[ \frac{y}{x} = \frac{39}{3} \][/tex]
Simplify:
[tex]\[ \frac{y}{x} = 13 \][/tex]
Hence:
[tex]\[ x = \frac{y}{13} \][/tex]
7. Verify the consistency with the exponential equation:
We have from the given problem:
[tex]\[ 2^x = 5 \][/tex]
Taking the logarithm with base 2 of both sides:
[tex]\[ x = \log_2{5} \][/tex]
8. Match both expressions for [tex]\(x\)[/tex]:
Set the two expressions for [tex]\(x\)[/tex] equal:
[tex]\[ \frac{y}{13} = \log_2{5} \][/tex]
Thus:
[tex]\[ y = 13 \log_2{5} \][/tex]
9. Determine [tex]\(z\)[/tex]:
Substitute [tex]\(y\)[/tex] back into the equation for [tex]\(z\)[/tex]:
[tex]\[ z = \frac{y}{40} = \frac{13 \log_2{5}}{40} \][/tex]
10. Substitute [tex]\(y\)[/tex] and [tex]\(z\)[/tex] back into the equation to be proven:
Verify:
[tex]\[ \frac{1}{z} = \frac{40}{13 \log_2{5}} \][/tex]
And on the other side:
[tex]\[ \frac{3}{x} + \frac{1}{y} = \frac{3}{\log_2{5}} + \frac{1}{13 \log_2{5}} \][/tex]
11. Simplify the right-hand side:
Find a common denominator and add:
[tex]\[ \frac{3}{\log_2{5}} + \frac{1}{13 \log_2{5}} = \frac{3 \cdot 13 + 1}{13 \log_2{5}} = \frac{40}{13 \log_2{5}} \][/tex]
12. Comparison:
Both sides of the equation are found to be:
[tex]\[ \frac{40}{13 \log_2{5}} = \frac{40}{13 \log_2{5}} \][/tex]
Thus, the equation is consistent and holds true:
[tex]\[ \frac{1}{z} = \frac{3}{x} + \frac{1}{y} \][/tex]
1. Expression for [tex]\(z\)[/tex] in terms of [tex]\(y\)[/tex]:
Given [tex]\(y = 40z\)[/tex], we can write:
[tex]\[ z = \frac{y}{40} \][/tex]
2. Substitute [tex]\(z\)[/tex] into the equation to be proven:
Substitute [tex]\(z\)[/tex] as [tex]\(\frac{y}{40}\)[/tex] in the equation:
[tex]\[ \frac{1}{z} = \frac{40}{y} \][/tex]
3. Rewrite the equation:
Now, we need to prove:
[tex]\[ \frac{40}{y} = \frac{3}{x} + \frac{1}{y} \][/tex]
4. Isolate the common denominator:
Multiply both sides of the equation by [tex]\(y\)[/tex] to clear the fraction:
[tex]\[ 40 = 3 \cdot \frac{y}{x} + 1 \][/tex]
Simplify:
[tex]\[ 40 = 3 \cdot \frac{y}{x} + 1 \][/tex]
5. Rearrange the terms:
Move 1 to the left side:
[tex]\[ 40 - 1 = 3 \cdot \frac{y}{x} \][/tex]
Simplify:
[tex]\[ 39 = 3 \cdot \frac{y}{x} \][/tex]
6. Isolate [tex]\(x\)[/tex]:
Rearrange to solve for [tex]\(x\)[/tex]:
[tex]\[ \frac{y}{x} = \frac{39}{3} \][/tex]
Simplify:
[tex]\[ \frac{y}{x} = 13 \][/tex]
Hence:
[tex]\[ x = \frac{y}{13} \][/tex]
7. Verify the consistency with the exponential equation:
We have from the given problem:
[tex]\[ 2^x = 5 \][/tex]
Taking the logarithm with base 2 of both sides:
[tex]\[ x = \log_2{5} \][/tex]
8. Match both expressions for [tex]\(x\)[/tex]:
Set the two expressions for [tex]\(x\)[/tex] equal:
[tex]\[ \frac{y}{13} = \log_2{5} \][/tex]
Thus:
[tex]\[ y = 13 \log_2{5} \][/tex]
9. Determine [tex]\(z\)[/tex]:
Substitute [tex]\(y\)[/tex] back into the equation for [tex]\(z\)[/tex]:
[tex]\[ z = \frac{y}{40} = \frac{13 \log_2{5}}{40} \][/tex]
10. Substitute [tex]\(y\)[/tex] and [tex]\(z\)[/tex] back into the equation to be proven:
Verify:
[tex]\[ \frac{1}{z} = \frac{40}{13 \log_2{5}} \][/tex]
And on the other side:
[tex]\[ \frac{3}{x} + \frac{1}{y} = \frac{3}{\log_2{5}} + \frac{1}{13 \log_2{5}} \][/tex]
11. Simplify the right-hand side:
Find a common denominator and add:
[tex]\[ \frac{3}{\log_2{5}} + \frac{1}{13 \log_2{5}} = \frac{3 \cdot 13 + 1}{13 \log_2{5}} = \frac{40}{13 \log_2{5}} \][/tex]
12. Comparison:
Both sides of the equation are found to be:
[tex]\[ \frac{40}{13 \log_2{5}} = \frac{40}{13 \log_2{5}} \][/tex]
Thus, the equation is consistent and holds true:
[tex]\[ \frac{1}{z} = \frac{3}{x} + \frac{1}{y} \][/tex]
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