Question

When (56)^a=(5.6)^b=(10)^c then prove 1/a=1/b+1/c

Answer 1

Solution :-

56^a = 5.6^b = 10^c

1) 56^a = 10^c

⇒ 56 = 10^(c/a) --- eq(1)

2) 5.6^b = 10^c

⇒ 5.6 = 10^(c/b)

Multiplying on both sides by 10

⇒ 5.6 * 10 = 10^(c/b) * 10¹

⇒ 56 = 10^(c/b + 1) ---- eq(2)

[ Because a^m * a^n = a^(m + n) ]

From eq(1) and eq(2)

⇒ 10^(c/a) = 10^(c/b + 1)

Since bases are equal we can equate powers

⇒ c/a = c/b + 1

⇒ c/a - c/b = 1

⇒ c(1/a - 1/b) = 1

⇒ 1/a - 1/b = 1/c

⇒ 1/a = 1/b + 1/c

Hence proved.

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Answer 2

Answer:

1 / a = 1 / b + 1 / c  [ Proved ]

Step-by-step explanation:

Given :

$\displaystyle {(56)^a=(5.6)^b=(10)^c}$

Taking log :

$\displaystyle{\log 56^a=\log 10^c}$

$\displaystyle{\log5.6^b=\log 10^c}$

Using log rule i.e if log aˣ = x log a

a log 56 = c log 10

= > log 56 = c / a

b log 5.6 = c log 10

log 5.6 = c / b

log 56 - log 10 = c /b

We have log 56 = c / a and know value of log 10 = 1

c / a - 1 = c / b

c ( 1 / a - 1 / b ) = 1

1 / a = 1 / b + 1 / c

Hence proved.