Question

2ᵃ = 5ᵇ = 10ⁿ.
Show that n = $\frac{ab}{a + b}$

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 = $\frac{c}{a}$
c = bln5; ln5 = $\frac{c}{b}$

Hence, c = n(ln2 + ln5) = n($\frac{c}{a}$ + $\frac{c}{b}$)
Factorise c outside on the right hand side.

c = cn($\frac{1}{a}$ + $\frac{1}{b}$)
1 = n($\frac{1}{a}$ + $\frac{1}{b}$)
$\frac{1}{n}$ = $\frac{1}{a}$ + $\frac{1}{b}$

$\frac{1}{n}$ = $\frac{a + b}{ab}$
and thus, n = $\frac{ab}{a + b}$

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 = $\sqrt[a]{c}$ = $c^{\frac{1}{a}}$

Then, 5 = $c^{\frac{1}{b}} $
and 10 = $c^{\frac{1}{n}} $

But, 10 = 5·2, so 10 = $c^{\frac{1}{b}} $·$c^{\frac{1}{a}} $
∴ $c^{\frac{1}{n}} $ = $c^{\frac{1}{b}} $·$c^{\frac{1}{a}} $

$\frac{1}{n}$ = $\frac{1}{a}$ + $\frac{1}{b}$
and n = $\frac{ab}{a + b}$