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If a^x=b^y=c^z then prove that:1/x+1/y+1/z=1​

User Sema
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Explanation:

To prove the given statement, we'll start by taking the natural logarithm of each side of the equation a^x = b^y = c^z. Applying the logarithmic property, we have:

ln(a^x) = ln(b^y) = ln(c^z)

Using the power rule of logarithms, we can rewrite this as:

x * ln(a) = y * ln(b) = z * ln(c)

Now, let's consider the equation x * ln(a) = y * ln(b) = z * ln(c). Since all three expressions are equal, we can write:

x * ln(a) = y * ln(b) = z * ln(c) = k, where k is some constant.

Now, let's solve for x:

x = k / ln(a)

Similarly, we can solve for y and z:

y = k / ln(b)

z = k / ln(c)

Now, let's express 1/x, 1/y, and 1/z in terms of k, ln(a), ln(b), and ln(c):

1/x = ln(a) / k

1/y = ln(b) / k

1/z = ln(c) / k

Adding these three expressions together, we get:

1/x + 1/y + 1/z = ln(a) / k + ln(b) / k + ln(c) / k

Combining the terms, we have:

1/x + 1/y + 1/z = (ln(a) + ln(b) + ln(c)) / k

Since x * ln(a) = y * ln(b) = z * ln(c) = k, we can rewrite this as:

1/x + 1/y + 1/z = ln(a^x * b^y * c^z) / k

Since a^x = b^y = c^z, we can further simplify this expression:

1/x + 1/y + 1/z = ln(a^x * b^y * c^z) / k = ln(a^x) / k = x * ln(a) / k = x / x = 1

Therefore, we have proved that if a^x = b^y = c^z, then 1/x + 1/y + 1/z = 1.

User Alex Sunder Singh
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