Answer

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**Hint:**Identical numbers in which each number is raised to the same power can be added or simply we can multiply one of the numbers with the number of times it is asked to be added.

**Complete step by step answer:**

Step 1: taking $2^x$ common from left hand side and we get

\[{2^x} + {2^x} + {2^x} = 192\]

\[{2^x}(1 + 1 + 1) = 192\]

Adding all the terms inside the bracket, we get

\[{2^x} \times 3 = 192\]

Step 2: since we do not know the value of x so it is not possible to multiply 3 with 2x

Therefore, taking 3 into right hand side in the denominator, we get

\[{2^x} \times 3 = 192\]

\[{2^x} = \dfrac{{192}}{3}\]

Simplifying further, we get

\[{2^x} = 64\]

The above equation simply means that the possible values of x such that the whole number in the left hand side, after simplification, can be equal to 64.

Step 3: now from factorization method we can write,

\[64 = 2 \times 2 \times 2 \times 2 \times 2 \times 2\]

Or

In terms of power we can write it as

\[64 = {2^6}\]

Now we have two equations, first is \[{2^x} = 64\] and second is \[64 = {2^6}\]

After comparing both equations we can write

\[{2^x} = 64\]

Now we can also write 64 as multiples of 2 i.e.

\[64 = {2^6}\]

Comparing both the equation, we get

\[ \Rightarrow \]\[{2^x} = {2^6}\]

Now, as we know that according to the law of exponent method

If\[{a^m} = {a^n}\], then \[m = n\]

So in equation \[{2^x} = {2^6}\] we have the same number on both sides and in which one is raised to power ‘x’ and other is raised to power ‘6’. Therefore, according to law of exponent we have

\[{2^x} = {2^6}\]

\[ \Rightarrow \]\[x = 6\]

Hence, that the value of \[x = 6\]

**Note:**Numbers with negative exponents obey the following laws of exponents,

\[{a^m} \times {a^n} = {a^{m + n}};{a^m} \div {a^n} = {a^{m - n}};{({a^m})^n} = {a^{mn}};{a^m} \times {b^m} = {(ab)^m};{a^0} = 1;\dfrac{{{a^m}}}{{{b^m}}} = {(\dfrac{a}{b})^m}\]

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