Question

How do you solve \[{(6b)^{\dfrac{1}{2}}} = {(8 - 2b)^{\dfrac{1}{2}}}\] ?

Answer 1

Hint: For this expression or kind of expression we have to solve for the powers first either take square root for both side of equation when power is of degree in both the side, or do squaring both side when root is given on both side of the equation, rest you can solve for the equation by solving the expression forward.

Complete step by step solution:
The given question is \[{(6b)^{\dfrac{1}{2}}} = {(8 - 2b)^{\dfrac{1}{2}}}\]. Here in the given question the power of the terms across the equals to sign are in the root, hence we have to do the square on both sides of the equation to replace the square root sign, and then we can easily simplify the expression for the result. On solving we get:
\[{(6b)^{\dfrac{1}{2}}} = {(8 - 2b)^{\dfrac{1}{2}}} \]
Squaring both side we get:
\[{\left( {6{b^{\dfrac{1}{2}}}} \right)^2} = {\left[ {{{\left( {8 - 2b} \right)}^{\dfrac{1}{2}}}} \right]^2} \\
\Rightarrow 6{b^{\dfrac{1}{2} \times 2}} = {\left( {8 - 2b} \right)^{\dfrac{1}{2} \times 2}} \\
\Rightarrow 6b = 8 - 2b \\ \]
Now we obtained the equation with maximum power of one, here we have to do simple maths to get the solution for the variable, on solving we get:
\[\Rightarrow 6b = 8 - 2b \\
\Rightarrow 6b + 2b = 8 \\
\Rightarrow 8b = 8 \\
\therefore b = \dfrac{8}{8} = 1 \]
Here we solved for the given variable and got the answer for the variable whose value is one for the given expression.

Note: For such expression in which the power of the terms are non singular that is any other power rhen one, then we have to do the first step of simplifying the power by the suitable means according to the question. Or you can go for any other mathematical method like taking “log” on both sides, but there you have to face problems when step involves an antilog, there you should have a log table.