
How does one solve ${9^{x - 1}} \times {81^{2x - 1}} = {27^{3x - 2}}$ ?
Answer
534.6k+ views
Hint: In the given question, we have been asked to find the value of ‘x’ and it is given that ${9^{x - 1}} \times {81^{2x - 1}} = {27^{3x - 2}}$ . To solve this question, we need to get ‘x’ on one side of the “equals” sign, and all the other numbers on the other side. To solve this equation for a given variable ‘x’, we have to undo the mathematical operations such as addition, subtraction, multiplication, and division that have been done to the variables.
Complete step by step answer:
It is given that, ${9^{x - 1}} \times {81^{2x - 1}} = {27^{3x - 2}}$.
We have to solve for $x$ .
Try to convert the given equation with the same bases .
We will get as follow ,
${9^{x - 1}} \times {81^{2x - 1}} = {27^{3x - 2}}$
$ \Rightarrow {3^{2(x - 1)}} \times {3^{4(2x - 1)}} = {3^{3(3x - 2)}}$
We can write this as ,
${3^{2x - 2}} \times {3^{8x - 4}} = {3^{9x - 6}}$
Now using the identity ${a^b} \times {a^c} = {a^{bc}}$ ,
${3^{2x - 2 + }}^{8x - 4} = {3^{9x - 6}} \\
\Rightarrow {3^{10x - 6}} = {3^{9x - 6}}$
Now we have same base , therefore we can equate the exponents as follow ,
$10x - 6 = 9x - 6$
Add $6$ both the side of the equation , we will get ,
$10x - 6 + 6 = 9x - 6 + 6 \\
\Rightarrow 10x = 9x$
Subtract $9x$ from both the side of the equation, we will get ,
$10x - 9x = 9x - 9x \\
\therefore x = 0 \\ $
Therefore, the value of $x$ is equal to $0$.
Additional information:
In the given question, mathematical operations such as addition, subtraction, multiplication and division is used. Use addition or subtraction properties of equality to gather variable terms on one side of the equation and constant on the other side of the equation. Use the multiplication or division properties of equality to form the coefficient of the variable term equivalent to one.
Note:The important thing to recollect about any equation is that the ‘equals’ sign represents a balance. What the sign says is that what’s on the left-hand side is strictly an equal to what’s on the right-hand side. It is the type of question where only mathematical operations such as addition, subtraction, multiplication and division is used.
Complete step by step answer:
It is given that, ${9^{x - 1}} \times {81^{2x - 1}} = {27^{3x - 2}}$.
We have to solve for $x$ .
Try to convert the given equation with the same bases .
We will get as follow ,
${9^{x - 1}} \times {81^{2x - 1}} = {27^{3x - 2}}$
$ \Rightarrow {3^{2(x - 1)}} \times {3^{4(2x - 1)}} = {3^{3(3x - 2)}}$
We can write this as ,
${3^{2x - 2}} \times {3^{8x - 4}} = {3^{9x - 6}}$
Now using the identity ${a^b} \times {a^c} = {a^{bc}}$ ,
${3^{2x - 2 + }}^{8x - 4} = {3^{9x - 6}} \\
\Rightarrow {3^{10x - 6}} = {3^{9x - 6}}$
Now we have same base , therefore we can equate the exponents as follow ,
$10x - 6 = 9x - 6$
Add $6$ both the side of the equation , we will get ,
$10x - 6 + 6 = 9x - 6 + 6 \\
\Rightarrow 10x = 9x$
Subtract $9x$ from both the side of the equation, we will get ,
$10x - 9x = 9x - 9x \\
\therefore x = 0 \\ $
Therefore, the value of $x$ is equal to $0$.
Additional information:
In the given question, mathematical operations such as addition, subtraction, multiplication and division is used. Use addition or subtraction properties of equality to gather variable terms on one side of the equation and constant on the other side of the equation. Use the multiplication or division properties of equality to form the coefficient of the variable term equivalent to one.
Note:The important thing to recollect about any equation is that the ‘equals’ sign represents a balance. What the sign says is that what’s on the left-hand side is strictly an equal to what’s on the right-hand side. It is the type of question where only mathematical operations such as addition, subtraction, multiplication and division is used.
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