Question

2ᵃ = 5ᵇ = 10ⁿ.
Show that n =
`(ab)/(a + b)`

Answer 1

There are two ways you can go about this: I'll explain both ways.

Solution 1: Using logarithmic properties
The first way is to use logarithmic properties.

We can take the natural logarithm to all three terms to utilise our exponents.

Hence, ln2ᵃ = ln5ᵇ = ln10ⁿ becomes:
aln2 = bln5 = nln10.

What's so neat about ln10 is that it's ln(5·2).
Using our logarithmic rule (log(ab) = log(a) + log(b),
we can rewrite it as aln2 = bln5 = n(ln2 + ln5)

Since it's equal (given to us), we can let it all equal to another variable "c".

So, c = aln2 = bln5 = n(ln2 + ln5) and the reason why we do this, is so that we may find ln2 and ln5 respectively.

c = aln2; ln2 =
`(c)/(a)`
c = bln5; ln5 =
`(c)/(b)`

Hence, c = n(ln2 + ln5) = n(
`(c)/(a)` +
`(c)/(b)`)
Factorise c outside on the right hand side.

c = cn(
`(1)/(a)` +
`(1)/(b)`)
1 = n(
`(1)/(a)` +
`(1)/(b)`)

`(1)/(n)` =
`(1)/(a)` +
`(1)/(b)`


`(1)/(n)` =
`(a + b)/(ab)`
and thus, n =
`(ab)/(a + b)`

Solution 2: Using exponent rules
In this solution, we'll be taking advantage of exponents.

So, let c = 2ᵃ = 5ᵇ = 10ⁿ
Since c = 2ᵃ, 2 =
`\sqrt[a]{c}` =
`c^{(1)/(a)}`

Then, 5 =
`c^{(1)/(b)}`
and 10 =
`c^{(1)/(n)}`

But, 10 = 5·2, so 10 =
`c^{(1)/(b)}`·
`c^{(1)/(a)}`
∴
`c^{(1)/(n)}` =
`c^{(1)/(b)}`·
`c^{(1)/(a)}`


`(1)/(n)` =
`(1)/(a)` +
`(1)/(b)`
and n =
`(ab)/(a + b)`