How do you solve \[{{\left( \dfrac{4}{3} \right)}^{x}}=\left( \dfrac{27}{64} \right)\]?
Answer
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Hint: Write \[\left( \dfrac{27}{64} \right)\] in the R.H.S. as exponent of \[\left( \dfrac{4}{3} \right)\] by using the formula: - \[\dfrac{1}{{{a}^{m}}}={{a}^{-m}}\]. Now, compare the bases on both the sides and equate the exponents to form a linear equation in x. Solve this equation for the value of x to get the answer.
Complete step by step answer:
Here, we have been provided with the exponential expression: - \[{{\left( \dfrac{4}{3} \right)}^{x}}=\left( \dfrac{27}{64} \right)\] and we are asked to solve it. That means we have to find the value of x.
Now, we can write \[\left( \dfrac{27}{64} \right)\] in the R.H.S. as exponential form with base \[\dfrac{3}{4}\]. Here, \[\dfrac{27}{64}=\dfrac{{{3}^{3}}}{{{4}^{3}}}\]. Using the identity \[\dfrac{{{a}^{m}}}{{{b}^{m}}}={{\left( \dfrac{a}{b} \right)}^{m}}\], we get,
\[\Rightarrow {{\left( \dfrac{4}{3} \right)}^{x}}={{\left( \dfrac{3}{4} \right)}^{3}}\]
Using the formula: - \[\dfrac{1}{{{a}^{m}}}={{a}^{-m}}\] in the R.H.S. to write \[{{\left( \dfrac{3}{4} \right)}^{3}}={{\left( \dfrac{4}{3} \right)}^{-3}}\], we get,
\[\Rightarrow {{\left( \dfrac{4}{3} \right)}^{x}}={{\left( \dfrac{4}{3} \right)}^{-3}}\]
As we can see that both the sides of the above exponential expression contains \[\left( \dfrac{4}{3} \right)\] as the base. So, we can equate the exponents by removing the base from both the sides. So, we have,
\[\Rightarrow x=-3\]
Hence, the value of x is -3.
Note: One may note that here we have used some basic formulas of the topic ‘exponents and powers’ to solve the question. You must remember some basic formulas such as: - \[{{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}\], \[\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}\] and \[{{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}\] because they are used with the help of logarithm. We can take log to the base \[\dfrac{4}{3}\] or \[\dfrac{3}{4}\], i.e., \[{{\log }_{\dfrac{4}{3}}}\]or \[{{\log }_{\dfrac{3}{4}}}\], both the sides and use the property \[{{\log }_{n}}n=1\] to get the answer. Here, n > 0 and \[n\ne 1\].
Complete step by step answer:
Here, we have been provided with the exponential expression: - \[{{\left( \dfrac{4}{3} \right)}^{x}}=\left( \dfrac{27}{64} \right)\] and we are asked to solve it. That means we have to find the value of x.
Now, we can write \[\left( \dfrac{27}{64} \right)\] in the R.H.S. as exponential form with base \[\dfrac{3}{4}\]. Here, \[\dfrac{27}{64}=\dfrac{{{3}^{3}}}{{{4}^{3}}}\]. Using the identity \[\dfrac{{{a}^{m}}}{{{b}^{m}}}={{\left( \dfrac{a}{b} \right)}^{m}}\], we get,
\[\Rightarrow {{\left( \dfrac{4}{3} \right)}^{x}}={{\left( \dfrac{3}{4} \right)}^{3}}\]
Using the formula: - \[\dfrac{1}{{{a}^{m}}}={{a}^{-m}}\] in the R.H.S. to write \[{{\left( \dfrac{3}{4} \right)}^{3}}={{\left( \dfrac{4}{3} \right)}^{-3}}\], we get,
\[\Rightarrow {{\left( \dfrac{4}{3} \right)}^{x}}={{\left( \dfrac{4}{3} \right)}^{-3}}\]
As we can see that both the sides of the above exponential expression contains \[\left( \dfrac{4}{3} \right)\] as the base. So, we can equate the exponents by removing the base from both the sides. So, we have,
\[\Rightarrow x=-3\]
Hence, the value of x is -3.
Note: One may note that here we have used some basic formulas of the topic ‘exponents and powers’ to solve the question. You must remember some basic formulas such as: - \[{{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}\], \[\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}\] and \[{{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}\] because they are used with the help of logarithm. We can take log to the base \[\dfrac{4}{3}\] or \[\dfrac{3}{4}\], i.e., \[{{\log }_{\dfrac{4}{3}}}\]or \[{{\log }_{\dfrac{3}{4}}}\], both the sides and use the property \[{{\log }_{n}}n=1\] to get the answer. Here, n > 0 and \[n\ne 1\].
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