Question

if 2^a=5^b=10^c prove that 1/a+1/b=1/c​

Answer 1

Let,

${2}^{a} = {5}^{b} = {10}^{c} = k$


Now,
The things we can get by using logarithm here is,
(a)

$log_{2}(k) = a \\ \frac{ log(k) }{ log(2) } = a \\ \\ \frac{ log(2) }{ log(k) } = \frac{1}{a}$


And,
(b)
$log_{5}(k) = b \\ \frac{ log(k) }{ log(5) } = b \\ \frac{ log(5) }{ log(k) } = \frac{1}{b}$


Similarly,

(c)
$log_{10}(k) = c \\ \frac{ log(k) }{ log(10) } = c \\ \frac{ log(10) }{ log(k) } = \frac{1}{c}$

Adding the first two equations gives,


$\frac{ log(2) + log(5) }{ log(k) }$
I'm gonna use an identity here where we have,
$log(a) + log(b) = log(ab)$


Using this,
we get,
$\frac{ log(10) }{ log(k) }$


Which is equal to (c).

Therefore,

$\frac{1}{a} + \frac{1}{b} = \frac{1}{c}$

Cheers!

Answer 2

Answer:

Let,

{2}^{a} = {5}^{b} = {10}^{c} = k2

a

=5

b

=10

c

=k

Now,

The things we can get by using logarithm here is,

(a)

\begin{gathered} log_{2}(k) = a \\ \frac{ log(k) }{ log(2) } = a \\ \\ \frac{ log(2) }{ log(k) } = \frac{1}{a} \end{gathered}

log

2

(k)=a

log(2)

log(k)

=a

log(k)

log(2)

=

a

1

And,

(b)

\begin{gathered} log_{5}(k) = b \\ \frac{ log(k) }{ log(5) } = b \\ \frac{ log(5) }{ log(k) } = \frac{1}{b} \end{gathered}

log

5

(k)=b

log(5)

log(k)

=b

log(k)

log(5)

=

b

1

Similarly,

(c)

\begin{gathered} log_{10}(k) = c \\ \frac{ log(k) }{ log(10) } = c \\ \frac{ log(10) }{ log(k) } = \frac{1}{c} \end{gathered}

log

10

(k)=c

log(10)

log(k)

=c

log(k)

log(10)

=

c

1

Adding the first two equations gives,

\begin{gathered} \frac{ log(2) + log(5) }{ log(k) } \\ \end{gathered}

log(k)

log(2)+log(5)

I'm gonna use an identity here where we have,

log(a) + log(b) = log(ab)log(a)+log(b)=log(ab)

Using this,

we get,

\begin{gathered} \frac{ log(10) }{ log(k) } \\ \end{gathered}

log(k)

log(10)

Which is equal to (c).

Therefore,

\begin{gathered} \frac{1}{a} + \frac{1}{b} = \frac{1}{c} \\ \end{gathered}

a

1

+

b

1

=

c

1