Solve the system of equations 2x-y=1; x+2y=8 graphically and find the coordinates of the points where corresponding lines intersect y-axis.
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Find the range which the logarithm can be defined
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$- \dfrac { 1 } { 2 } \log _{ 10 } { \left( x \right) }$
Simplify the expression
$\log _{ 10 } { \left( \dfrac { \color{#FF6800}{ 1 } } { \sqrt{ x } } \right) }$
$ $ Since the logarithm of antilogarithm numbers and numerator is 1 as the fraction, add minus to the logarithm and take reciprocal to antilogarithm numbers $ $
$\color{#FF6800}{ - } \log _{ 10 } { \left( \sqrt{ x } \right) }$
$- \log _{ 10 } { \left( \sqrt{ \color{#FF6800}{ x } } \right) }$
$ $ Convert the square root of the antilogarithm number of the logarithm to the power $ $
$- \log _{ 10 } { \left( \color{#FF6800}{ x } ^ { \color{#FF6800}{ \frac { \color{#FF6800}{ 1 } } { \color{#FF6800}{ 2 } } } } \right) }$
$- \log _{ \color{#FF6800}{ 10 } } { \left( \color{#FF6800}{ x } ^ { \color{#FF6800}{ \frac { \color{#FF6800}{ 1 } } { \color{#FF6800}{ 2 } } } } \right) }$
$ $ Simplify the expression using $ \log_{a}{b^{x}}=x\times\log_{a}{b}$
$- \left ( \color{#FF6800}{ \dfrac { \color{#FF6800}{ 1 } } { \color{#FF6800}{ 2 } } } \log _{ \color{#FF6800}{ 10 } } { \left( \color{#FF6800}{ x } \right) } \right )$
$\color{#FF6800}{ - } \left ( \color{#FF6800}{ \dfrac { \color{#FF6800}{ 1 } } { \color{#FF6800}{ 2 } } } \log _{ \color{#FF6800}{ 10 } } { \left( \color{#FF6800}{ x } \right) } \right )$
$ $ Get rid of unnecessary parentheses $ $
$\color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 1 } } { \color{#FF6800}{ 2 } } } \log _{ \color{#FF6800}{ 10 } } { \left( \color{#FF6800}{ x } \right) }$
$x > 0$
$ $ Find the range of $ x $ where the logarithm is defined $ $
$\log _{ \color{#FF6800}{ 10 } } { \left( \color{#FF6800}{ \dfrac { \color{#FF6800}{ 1 } } { \sqrt{ \color{#FF6800}{ x } } } } \right) }$
$ $ Find the interval of $ x $ so that the antilogarithm number of logarithm is a positive number $ $
$\color{#FF6800}{ \dfrac { \color{#FF6800}{ 1 } } { \sqrt{ \color{#FF6800}{ x } } } } > \color{#FF6800}{ 0 }$
$\color{#FF6800}{ \dfrac { \color{#FF6800}{ 1 } } { \sqrt{ \color{#FF6800}{ x } } } } > \color{#FF6800}{ 0 }$
$ $ Solve the inequality $ $
$\color{#FF6800}{ x } > \color{#FF6800}{ 0 }$
Solution search results
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In the following problem $a,b,$ b,and c represent REAL NUMBERS. The derivative of $g\left(x\right)=log _{a}\left(bx\right)+cx^{2}is$ $○$ $g^{1}\left(x\right)=\right)dfrac\left(b\right)log _{-}a\left(bx\right)\right)\left(N1n$ $a\right)+2cx1\right)$ $○$ $\left(g\left(x\right)=\right)dtracb$ $loga\left(bx\right)+2c\times \right)Nln$ a}\) $○$ $g^{'}\left(x\right)=\left(dfraC\left(a\right)log _{-}a\left(bx\right)\right)\left(ln$ $\right)+2c\times 1\right)\right)$ $○$ $g^{1}\left(x\right)=\left(dfrac\left(b$ $log _{-}a\left(bx\right)\right)\left(ln$ $a1\right)$ $○$ $\left(\left(g^{1}\left(x\right)=b$ $log _{-}a\left(bx\right)+2cx1\right)$ $○$ $g\left(x\right)=\left(dfrac\left(b$ $log _{-}a\left(bx\right)\right)\left(1ln$ $a|+2c1\right)$
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