Question

How do you write $''4x$ (times) $4x$ (times) $4x''$ in exponential form?

Answer 1

Hint: This problem can be solved using law of indices.
${{a}^{m}} \times {{a}^{n}}={{a}^{m+n}}$ This is first law of indices which states that when the two terms have the same base and the terms have to multiplied together then their indices are added.
${{\left( ab \right)}^{n}}={{a}^{n}}.{{b}^{n}}$ In this formula the powers of the terms are separated and given to each. These two formulas will help you in solving problems.

Complete step by step solution:
Here, we have to write $''4x$ (times) $4x$ (times) $4x''$ in exponential form.
That is we have to write $4x.4x.4x$ in exponential form.
Now, $4x$ can be written as also ${{\left( 4x \right)}^{2}}$
Therefore, $4x={{\left( 4x \right)}^{1}}$
So, we have.
$\Rightarrow \left( 4x.4x.4x \right)={{\left( 4x \right)}^{1}}{{\left( 4x \right)}^{1}}{{\left( 4x \right)}^{1}}$
We know that, when multiplying the exponents having the same base value, we have to simply add their power and take base value as common. That is we can write above equation as,
$\left( 4x.4x.4x \right)={{\left( 4x \right)}^{1}}{{\left( 4x \right)}^{1}}{{\left( 4x \right)}^{1}}$
Now,
Taking base as common and adding the power then equation will be,
$\left( 4x.4x.4x \right)={{\left( 4x \right)}^{\left( 1+1+1 \right)}}$
$\Rightarrow \left( 4x.4x.4x \right)={{\left( 4x \right)}^{3}}$
Now, separating the power of term ${{\left( 4x \right)}^{3}}$ then we have,
$\left( 4x.4x.4x \right)={{4}^{3}}{{x}^{3}}$
As we know that,
${{4}^{3}}=4\times 4\times 4=64$ and keeping ${{x}^{3}}$ as it is.
Therefore we have the equation as.
$\left( 4x.4x.4x \right)=64{{x}^{3}}$
Hence, $''4x$ $\times$ $4x$ $\times$ $4x''$ in exponential form can be written as $64{{x}^{3}}$
The term exponential means the number of times a number multiplied by itself. Here, we multiply the numbers and convert into the simplest and short form of exponential. Use laws of indices for solving problems.

Note: We have to take precaution when multiplying the exponents term with the same base value. We have not to multiply the power of bases. We have to take base value as common and perform addition of their powers only.