What is log_{10}(2x)log_{10}(5)-1-2log_{10}(5/2) ?

The solution to log_{10}(2x)log_{10}(5)-1-2log_{10}(5/2) is log_{10}((4(2x)^{log_{10}(5)})/(25))-1

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simplify log_{10}(2x)log_{10}(5)-1-2log_{10}(5/2)

simplify

lo g_{10} (2 x) lo g_{10} (5) - 1 - 2 lo g_{10} (\frac{5}{2})

lo g_{10} (\frac{4 ( 2 x ) ^{l o g_{10} (5)}}{25}) - 1

Solution steps

lo g_{10} (2 x) lo g_{10} (5) - 1 - 2 lo g_{10} (\frac{5}{2})

Apply log rule:

a lo g_{c} (b) = lo g_{c} (b^{a})

lo g_{10} (5) lo g_{10} (2 x) = lo g_{10} ((2 x)^{l o g_{10} (5)})

= lo g_{10} ((2 x)^{l o g_{10} (5)}) - 1 - 2 lo g_{10} (\frac{5}{2})

Apply log rule:

a lo g_{c} (b) = lo g_{c} (b^{a})

2 lo g_{10} (\frac{5}{2}) = lo g_{10} ((\frac{5}{2})^{2})

= lo g_{10} ((2 x)^{l o g_{10} (5)}) - 1 - lo g_{10} ((\frac{5}{2})^{2})

Apply log rule:

lo g_{c} (a) - lo g_{c} (b) = lo g_{c} (\frac{a}{b})

lo g_{10} ((2 x)^{l o g_{10} (5)}) - lo g_{10} ((\frac{5}{2})^{2}) = lo g_{10} (\frac{( 2 x ) ^{l o g_{10} (5)}}{( \frac{5}{2} ) ^{2}})

= lo g_{10} (\frac{( 2 x ) ^{l o g_{10} (5)}}{( \frac{5}{2} ) ^{2}}) - 1

Simplify

\frac{( 2 x ) ^{l o g_{10} (5)}}{( \frac{5}{2} ) ^{2}} : \frac{4 ( 2 x ) ^{l o g_{10} (5)}}{25}

= lo g_{10} (\frac{4 ( 2 x ) ^{l o g_{10} (5)}}{25}) - 1

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What is log_{10}(2x)log_{10}(5)-1-2log_{10}(5/2) ?
The solution to log_{10}(2x)log_{10}(5)-1-2log_{10}(5/2) is log_{10}((4(2x)^{log_{10}(5)})/(25))-1