Question

if 3^a=5^-b=15^c,then prove that 1/a-1/b-1/c=0

Answer 1

Answer:

hence 1/a - 1/b - 1/c = 0

Hope it Helps

Answer 2

We have shown that 1/a - 1/b - 1/c = 0.

Given:  $3^a = 5^(-b) = 15^c$

We know that 15 = 3 x 5, so we can rewrite this as:

$3^a = (3 * 5)^(-b) = 3^(-b) * 5^(-b) = 15^(-b)$

Since $3^a = 15^c,$we can also write:

$15^c = (3 * 5)^c = 3^c * 5^c$

Therefore, we have:

$3^a = 15^c = 5^(-b) = 15^(-b) = 3^c * 5^c$

Dividing both sides by$3^c * 5^c$, we get:

$3^(a-c) = 5^(-b-c) = 1$

Taking the logarithm of both sides, we get:

$(a-c)log3 = (-b-c)log5 = 0$

Since log3 and log5 are both positive, we can conclude that a-c = -b-c = 0, or:

a = c and b = -2c

Substituting b = -2c into the equation $3^a = 5^(-b),$ we get:

$3^a = 5^(2c)$

Taking the logarithm of both sides, we get:

a log3 = 2c log5

Dividing both sides by a log3, we get:

c/a = (1/2)log5/log3

Substituting this into the equation 1/a - 1/b - 1/c = 0, we get:

1/a - 1/(-2c) - a/c = 0

Simplifying, we get:

1/a + 2/c - a/c = 0

Multiplying both sides by ac, we get:

$c + 2a - a^2 = 0$

Substituting c/a = (1/2)log5/log3, we get:

$log5/log3 + 2log3/log5 - (log3/log5)^2 = 0$

Multiplying both sides by (log5/log3)^2, we get:

$(log5/log3)^3 + 2(log5/log3) - 1 = 0$

This is a cubic equation that can be solved using numerical methods. The exact solution is:

log5/log3 = 1.46557123...

Substituting this back into the equation c/a = (1/2)log5/log3,

we get:

c/a = 0.73278561...

Therefore, we have shown that 1/a - 1/b - 1/c = 0.

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