Question

NEED IT URGENT...PLZ HELP​

Answer 1

Qᴜᴇsᴛɪᴏɴ :-

• If 2^a = 3^b = 6^c ,, Then Show that c = ab/(a + b).

Sᴏʟᴜᴛɪᴏɴ :-

Let us Assume That, 2^a = 3^b = 6^c = k (where k is any Constant Number).

So,

→ when 2^a = k

→ 2 = k^(1/a)

Similarly,

→ when 3^b = k

→ 3 = k^(1/b)

And,

→ when 6^c = k

→ 6 = k^(1/c)

Now, we can write 6 as Multiple of 2 and 3.

Putting values of 2 & 3 in 6 we get :-

→ 2 * 3 = 6

→ k^(1/a) * k^(1/b) = k^(1/c)

using a^b * a^c = a^(b + c) in LHS now,

→ k^(1/a + 1/b) = k^(1/c)

Now, we know that, when base is same , than power are Equal, or if a^b = a^c , Than b = c,

So,

→ (1/a + 1/b) = (1/c)

→ (a + b) /ab = 1/c

Or,

→ c = ab / (a + b) (Hence, Proved).

Answer 2

Step-by-step explanation:

let 2^a=3^b=6^c= k

2 =k^1/a

3=k^1/b

6=k^1/c

2×3=6

k^1/a ×k^1/b=k^1/c

k^(1/a+1/b) =k^1/c

1/a + 1/b = 1/c

b + a / ab = 1/c

hence

c = ab /a + b