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umka2103 [35]
3 years ago
10

2ᵃ = 5ᵇ = 10ⁿ. Show that n =

e=" \frac{ab}{a + b} " alt=" \frac{ab}{a + b} " align="absmiddle" class="latex-formula">
Mathematics
1 answer:
11Alexandr11 [23.1K]3 years ago
4 0
There are two ways you can go about this: I'll explain both ways.
<span>
</span><span>Solution 1: Using logarithmic properties
</span>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}

<span>Solution 2: Using exponent rules
</span>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}
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