Math, asked by manassharm03p8hvbn, 1 year ago

if (a^m)^n=a^m^n then express m in terms of n

Answers

Answered by Vickuvikas
35

Answer:

Step-by-step explanation:

I hope my ans is very helpful for you

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Answered by bharathparasad577
1

Answer:

Concept:

There are various exponent laws in mathematics. Many mathematical problems that involve repeated multiplication operations can be solved using all of the exponentiation principles. The principles of exponents make multiplication and division processes simpler and make problem-solving simple.

Step-by-step explanation:

Method 1:

            $$\begin{aligned}&\left(a^{\mathrm{m}}\right)^{\mathrm{n}}=a^{\mathrm{m}^{\mathrm{n}}} \\&\Rightarrow a^{\mathrm{mn}}=a^{\mathrm{m}^{\mathrm{n}}}\end{aligned}\\Comparing \ exponents \ of \ both \ the \ sides, \ we \ get\begin{aligned}&\mathrm{mn}=\mathrm{m}^{\mathrm{n}} \\&\Rightarrow \frac{\mathrm{m}^{\mathrm{n}}}{\mathrm{m}}=\mathrm{n} \\&\Rightarrow \mathrm{m}^{(\mathrm{n}-1)}=\mathrm{n} \\&\therefore \mathrm{m}=\mathrm{n}^{\frac{1}{(\mathrm{n}-1)}}\end{aligned}$$

Method 2:

                $$\mathrm{m} \times \mathrm{n}=\mathrm{m}^{\mathrm{n}}$$

Taking logarithm to the base 'm' on both sides

               $$\begin{aligned}&\log _{\mathrm{m}} \mathrm{mn}=\log _{\mathrm{m}} \mathrm{m}^{\mathrm{n}} \\&\log _{\mathrm{m}} \mathrm{m}+\log _{\mathrm{m}} \mathrm{n}=\mathrm{n} \log _{\mathrm{m}} \mathrm{m} \\&1+\log _{\mathrm{m}} \mathrm{n}=\mathrm{n} \\&\log _{\mathrm{m}} \mathrm{n}=\mathrm{n}-1 \\&\mathrm{~m}^{\mathrm{n}-1}=\mathrm{n} \\&\therefore \mathrm{m}=\mathrm{n}^\frac{1}{\mathrm{n}-1}\end{aligned}$$

#SPJ2

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