
If we have a logarithmic inequality \[{{\log }_{0.3}}\left( x-1 \right)< {{\log }_{0.09}}\left( x-1 \right)\], then find the interval in which x lies.
A). x > 2
B). x < 2
C). x > -2
D). none of these
Answer
585k+ views
Hint: At first use the fact that in \[{{\log }_{a}}b\], b is always positive. So, $x – 1 > 0$ or $x > 1.$ After that solve it further then use the fact that if \[{{\log }_{a}}b< 0\] and a lies between 0 and 1 then we can write it as $b > 1$ and hence get the value of x.
Complete step-by-step solution:
In the question we are given an equation \[{{\log }_{0.3}}\left( x-1 \right)<{{\log }_{0.09}}\left( x-1 \right)\] and we have to find values for x for which the given equation satisfies according to the given options.
So the equation in the question is,
\[{{\log }_{0.3}}\left( x-1 \right)< {{\log }_{0.09}}\left( x-1 \right)\]
In the term \[{{\log }_{a}}b\] the expression or term b should be positive according to the definition of logarithm.
So, x – 1 should be positive or greater than 0.
Hence, $x – 1 > 0$ or $x > 1.$
Now we will use the identity that, \[{{\log }_{{{a}^{2}}}}b=\dfrac{1}{2}{{\log }_{a}}b\].
So, the given equation,
\[{{\log }_{0.3}}\left( x-1 \right)< {{\log }_{0.09}}\left( x-1 \right)\] can be written as,
\[{{\log }_{0.3}}\left( x-1 \right)< {{\log }_{{{\left( 0.3 \right)}^{2}}}}\left( x-1 \right)\]
Or, \[{{\log }_{0.3}}\left( x-1 \right)< \dfrac{1}{2}{{\log }_{0.3}}\left( x-1 \right)\].
Now multiplying by 2 on both the sides so we get,
\[2{{\log }_{0.3}}\left( x-1 \right)< {{\log }_{0.3}}\left( x-1 \right)\]
So, we can write it as,
\[{{\log }_{0.3}}\left( x-1 \right)< 0\]
Now if \[{{\log }_{a}}b< 0\] and a is greater that 0 but less than 1 then we can write it as b > 1.
Now applying this we can write, \[{{\log }_{0.3}}\left( x-1 \right)< 0\].
As, $x – 1 > 1$
Or, $x > 2$
Hence, the correct option is (a).
Note: Generally if an inequation is given let’s say \[{{\log }_{a}}b< c\], where a, b, c are constants. If a is great equation than 1 then \[b< {{a}^{c}}\]. This is the most common mistake students generally do while solving inequalities related to logarithms.
Complete step-by-step solution:
In the question we are given an equation \[{{\log }_{0.3}}\left( x-1 \right)<{{\log }_{0.09}}\left( x-1 \right)\] and we have to find values for x for which the given equation satisfies according to the given options.
So the equation in the question is,
\[{{\log }_{0.3}}\left( x-1 \right)< {{\log }_{0.09}}\left( x-1 \right)\]
In the term \[{{\log }_{a}}b\] the expression or term b should be positive according to the definition of logarithm.
So, x – 1 should be positive or greater than 0.
Hence, $x – 1 > 0$ or $x > 1.$
Now we will use the identity that, \[{{\log }_{{{a}^{2}}}}b=\dfrac{1}{2}{{\log }_{a}}b\].
So, the given equation,
\[{{\log }_{0.3}}\left( x-1 \right)< {{\log }_{0.09}}\left( x-1 \right)\] can be written as,
\[{{\log }_{0.3}}\left( x-1 \right)< {{\log }_{{{\left( 0.3 \right)}^{2}}}}\left( x-1 \right)\]
Or, \[{{\log }_{0.3}}\left( x-1 \right)< \dfrac{1}{2}{{\log }_{0.3}}\left( x-1 \right)\].
Now multiplying by 2 on both the sides so we get,
\[2{{\log }_{0.3}}\left( x-1 \right)< {{\log }_{0.3}}\left( x-1 \right)\]
So, we can write it as,
\[{{\log }_{0.3}}\left( x-1 \right)< 0\]
Now if \[{{\log }_{a}}b< 0\] and a is greater that 0 but less than 1 then we can write it as b > 1.
Now applying this we can write, \[{{\log }_{0.3}}\left( x-1 \right)< 0\].
As, $x – 1 > 1$
Or, $x > 2$
Hence, the correct option is (a).
Note: Generally if an inequation is given let’s say \[{{\log }_{a}}b< c\], where a, b, c are constants. If a is great equation than 1 then \[b< {{a}^{c}}\]. This is the most common mistake students generally do while solving inequalities related to logarithms.
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