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Use the Change-Of-Base Formula to Rewrite the Logarithm as a Ratio

Question 36

Multiple Choice

Use the change-of-base formula to rewrite the logarithm as a ratio of logarithms.Then use a graphing utility to graph the ratio.? f(x) =log⁡15xf ( x ) = \log _ { \frac { 1 } { 5 } } x ?


A) f(x) =log⁡15log⁡x=ln⁡15ln⁡xf ( x ) = \frac { \log \frac { 1 } { 5 } } { \log x } = \frac { \ln \frac { 1 } { 5 } } { \ln x }  Use the change-of-base formula to rewrite the logarithm as a ratio of logarithms.Then use a graphing utility to graph the ratio.?  f ( x )  = \log _ { \frac { 1 } { 5 } } x  ? A)   f ( x )  = \frac { \log \frac { 1 } { 5 } } { \log x } = \frac { \ln \frac { 1 } { 5 } } { \ln x }    B)   f ( x )  = \frac { \log x } { \log \frac { 1 } { 5 } } = \frac { \ln x } { \ln \frac { 1 } { 5 } }    C)   f ( x )  = \log x + \log \frac { 1 } { 5 } = \ln x + \ln \frac { 1 } { 5 }    D)   f ( x )  = \log \frac { \frac { 1 } { 5 } } { x } = \ln \frac { \frac { 1 } { 5 } } { x }    E)   f ( x )  = \log \frac { x } { \frac { 1 } { 5 } } = \ln \frac { x } { \frac { 1 } { 5 } }
B) f(x) =log⁡xlog⁡15=ln⁡xln⁡15f ( x ) = \frac { \log x } { \log \frac { 1 } { 5 } } = \frac { \ln x } { \ln \frac { 1 } { 5 } }  Use the change-of-base formula to rewrite the logarithm as a ratio of logarithms.Then use a graphing utility to graph the ratio.?  f ( x )  = \log _ { \frac { 1 } { 5 } } x  ? A)   f ( x )  = \frac { \log \frac { 1 } { 5 } } { \log x } = \frac { \ln \frac { 1 } { 5 } } { \ln x }    B)   f ( x )  = \frac { \log x } { \log \frac { 1 } { 5 } } = \frac { \ln x } { \ln \frac { 1 } { 5 } }    C)   f ( x )  = \log x + \log \frac { 1 } { 5 } = \ln x + \ln \frac { 1 } { 5 }    D)   f ( x )  = \log \frac { \frac { 1 } { 5 } } { x } = \ln \frac { \frac { 1 } { 5 } } { x }    E)   f ( x )  = \log \frac { x } { \frac { 1 } { 5 } } = \ln \frac { x } { \frac { 1 } { 5 } }
C) f(x) =log⁡x+log⁡15=ln⁡x+ln⁡15f ( x ) = \log x + \log \frac { 1 } { 5 } = \ln x + \ln \frac { 1 } { 5 }  Use the change-of-base formula to rewrite the logarithm as a ratio of logarithms.Then use a graphing utility to graph the ratio.?  f ( x )  = \log _ { \frac { 1 } { 5 } } x  ? A)   f ( x )  = \frac { \log \frac { 1 } { 5 } } { \log x } = \frac { \ln \frac { 1 } { 5 } } { \ln x }    B)   f ( x )  = \frac { \log x } { \log \frac { 1 } { 5 } } = \frac { \ln x } { \ln \frac { 1 } { 5 } }    C)   f ( x )  = \log x + \log \frac { 1 } { 5 } = \ln x + \ln \frac { 1 } { 5 }    D)   f ( x )  = \log \frac { \frac { 1 } { 5 } } { x } = \ln \frac { \frac { 1 } { 5 } } { x }    E)   f ( x )  = \log \frac { x } { \frac { 1 } { 5 } } = \ln \frac { x } { \frac { 1 } { 5 } }
D) f(x) =log⁡15x=ln⁡15xf ( x ) = \log \frac { \frac { 1 } { 5 } } { x } = \ln \frac { \frac { 1 } { 5 } } { x }  Use the change-of-base formula to rewrite the logarithm as a ratio of logarithms.Then use a graphing utility to graph the ratio.?  f ( x )  = \log _ { \frac { 1 } { 5 } } x  ? A)   f ( x )  = \frac { \log \frac { 1 } { 5 } } { \log x } = \frac { \ln \frac { 1 } { 5 } } { \ln x }    B)   f ( x )  = \frac { \log x } { \log \frac { 1 } { 5 } } = \frac { \ln x } { \ln \frac { 1 } { 5 } }    C)   f ( x )  = \log x + \log \frac { 1 } { 5 } = \ln x + \ln \frac { 1 } { 5 }    D)   f ( x )  = \log \frac { \frac { 1 } { 5 } } { x } = \ln \frac { \frac { 1 } { 5 } } { x }    E)   f ( x )  = \log \frac { x } { \frac { 1 } { 5 } } = \ln \frac { x } { \frac { 1 } { 5 } }
E) f(x) =log⁡x15=ln⁡x15f ( x ) = \log \frac { x } { \frac { 1 } { 5 } } = \ln \frac { x } { \frac { 1 } { 5 } }  Use the change-of-base formula to rewrite the logarithm as a ratio of logarithms.Then use a graphing utility to graph the ratio.?  f ( x )  = \log _ { \frac { 1 } { 5 } } x  ? A)   f ( x )  = \frac { \log \frac { 1 } { 5 } } { \log x } = \frac { \ln \frac { 1 } { 5 } } { \ln x }    B)   f ( x )  = \frac { \log x } { \log \frac { 1 } { 5 } } = \frac { \ln x } { \ln \frac { 1 } { 5 } }    C)   f ( x )  = \log x + \log \frac { 1 } { 5 } = \ln x + \ln \frac { 1 } { 5 }    D)   f ( x )  = \log \frac { \frac { 1 } { 5 } } { x } = \ln \frac { \frac { 1 } { 5 } } { x }    E)   f ( x )  = \log \frac { x } { \frac { 1 } { 5 } } = \ln \frac { x } { \frac { 1 } { 5 } }

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