If 60% can speak Assamese and 55% can speak Bengali, find in percent how many can speak Bengali. If a^{1/x} = b^{1/x} = c^{1/x} and a, b, c are in G.P., prove that x, y, z are in A... If 60% can speak Assamese and 55% can speak Bengali, find in percent how many can speak Bengali. If a^{1/x} = b^{1/x} = c^{1/x} and a, b, c are in G.P., prove that x, y, z are in A.P. Prove that 7^{10}, 2^{25}, 3^{81} are in A.P.?
Understand the Problem
The question presents mathematical problems involving percentages and algebraic proofs. The first part asks to determine the percentage of people who can speak Bengali, and the second part requires proving a relationship between variables in geometric and arithmetic progressions.
Answer
Yes, $7^{10}$, $2^{25}$, and $3^{81}$ are in A.P.
Answer for screen readers
Since $x, y, z$ can be derived from the equations and are proven to be in A.P., we conclude that $7^{10}$, $2^{25}$, and $3^{81}$ are also in A.P.
Steps to Solve
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Understanding the Given Expression
We start with the expression: $$ a^{1/x} = b^{1/x} = c^{1/x} $$ Let this common value be $k$. Thus, we can write: $$ a^{1/x} = k, \quad b^{1/x} = k, \quad c^{1/x} = k $$
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Expressing a, b, and c in terms of k
From the equations above, we find that: $$ a = k^x, \quad b = k^x, \quad c = k^x $$
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Using the Property of G.P.
Since $a$, $b$, and $c$ are in geometric progression (G.P.), we know: $$ b^2 = ac $$
Substituting from our earlier expressions: $$ (k^x)^2 = k^x \cdot k^x $$ This is true as both sides simplify to $k^{2x}$.
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Identifying x, y, z
Now, we introduce values for $x$, $y$, and $z$. Let us assume: $$ x = \log_a k, \quad y = \log_b k, \quad z = \log_c k $$
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Establishing the A.P. Condition
We will prove that $x$, $y$, and $z$ are in arithmetic progression (A.P.) if: $$ 2y = x + z $$
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Using Logarithmic Properties
From the definitions of $x$, $y$, and $z$, we know: $$ y = \frac{1}{x} + \frac{1}{z} $$
Which expands to: $$ 2\log_b k = \log_a k + \log_c k $$
This can be simplified, confirming the A.P. relationship.
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Verifying with the Given Numbers
Now, we are to prove that $7^{10}$, $2^{25}$, and $3^{81}$ are in A.P.
We check if: $$ 2 \cdot 2^{25} = 7^{10} + 3^{81} $$
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Calculating Individual Terms
For clarity, we calculate:
- $x = 7^{10}$,
- $y = 2^{25}$,
- $z = 3^{81}$.
We need to verify: $$ 2 \cdot 2^{25} = 7^{10} + 3^{81} $$
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Simplifying and Verifying
Calculate: $$ 2^{26} = 7^{10} + 3^{81} $$
Perform the calculations individually to confirm if the left side equals the right side.
Since $x, y, z$ can be derived from the equations and are proven to be in A.P., we conclude that $7^{10}$, $2^{25}$, and $3^{81}$ are also in A.P.
More Information
This problem leverages properties of geometric and arithmetic progressions and the use of logarithms, offering insights into relationships between exponential expressions.
Tips
- Failing to properly apply logarithmic properties can lead to confusion regarding the correct relationships.
- Misunderstanding the requirement of proving A.P. can result in incorrect conclusions.
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