# Find the sum of all the values of x satisfying ${\log _{10}}\left( {x + 9} \right) + 2{\log_ {10}}\sqrt {2x - 1} = 1$.

Last updated date: 26th Mar 2023

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Answer

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Hint: In this question we have to find the sum of all values of x, so using the property of logarithm we know that ${\log _{10}}10 = 1$ . This property will help you simplify things up and will eventually help you reach the right answer.

Complete step-by-step answer:

We have been given the expression ${\log _{10}}\left( {x + 9} \right) + 2{\log _{10}}\sqrt {2x - 1} = 1$.

The RHS has been given as 1.

Now as we know that ${\log _{10}}10 = 1$.

So, the RHS is equal to ${\log _{10}}10$.

Now, the given expression becomes ${\log _{10}}\left( {x + 9} \right) + 2{\log _{10}}\sqrt {2x - 1} = {\log _{10}}10$

Using the logarithmic property, $a\log x = \log {x^a}$ we get,

$ \Rightarrow {\log _{10}}\left( {x + 9} \right) + {\log _{10}}{\left( {\sqrt {2x - 1} } \right)^2} = {\log _{10}}10$

$ \Rightarrow {\log _{10}}\left( {x + 9} \right) + {\log _{10}}\left( {2x - 1} \right) = {\log _{10}}10$

Now when we use the property, $\log a + \log b = \log ab$ we get,

$ \Rightarrow {\log _{10}}\left( {x + 9} \right)\left( {2x - 1} \right) = {\log _{10}}10$

Comparing the LHS and RHS we get,

$ \Rightarrow \left( {x + 9} \right)\left( {2x - 1} \right) = 10$

$ \Rightarrow {x^2} + 17x - 9 = 0$

Now, we don’t need to solve this equation, we just need to find the sum of all values of x. We also know that the negative of the coefficient of x is equal to the sum of roots of an equation.

The coefficient of x is 17.

And hence the sum of all values of x is -17.

Note: Whenever we face such types of problems the key point to remember is that we need to have a good grasp over the logarithmic identities, some of them have been mentioned above. Along with this we should also be familiar with quadratic equations and its properties. These identities help you in simplification and getting on the right track to reach the answer.

Complete step-by-step answer:

We have been given the expression ${\log _{10}}\left( {x + 9} \right) + 2{\log _{10}}\sqrt {2x - 1} = 1$.

The RHS has been given as 1.

Now as we know that ${\log _{10}}10 = 1$.

So, the RHS is equal to ${\log _{10}}10$.

Now, the given expression becomes ${\log _{10}}\left( {x + 9} \right) + 2{\log _{10}}\sqrt {2x - 1} = {\log _{10}}10$

Using the logarithmic property, $a\log x = \log {x^a}$ we get,

$ \Rightarrow {\log _{10}}\left( {x + 9} \right) + {\log _{10}}{\left( {\sqrt {2x - 1} } \right)^2} = {\log _{10}}10$

$ \Rightarrow {\log _{10}}\left( {x + 9} \right) + {\log _{10}}\left( {2x - 1} \right) = {\log _{10}}10$

Now when we use the property, $\log a + \log b = \log ab$ we get,

$ \Rightarrow {\log _{10}}\left( {x + 9} \right)\left( {2x - 1} \right) = {\log _{10}}10$

Comparing the LHS and RHS we get,

$ \Rightarrow \left( {x + 9} \right)\left( {2x - 1} \right) = 10$

$ \Rightarrow {x^2} + 17x - 9 = 0$

Now, we don’t need to solve this equation, we just need to find the sum of all values of x. We also know that the negative of the coefficient of x is equal to the sum of roots of an equation.

The coefficient of x is 17.

And hence the sum of all values of x is -17.

Note: Whenever we face such types of problems the key point to remember is that we need to have a good grasp over the logarithmic identities, some of them have been mentioned above. Along with this we should also be familiar with quadratic equations and its properties. These identities help you in simplification and getting on the right track to reach the answer.

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