What is log_{4}(x-1)+log_{4}(2x-5) ?

The solution to log_{4}(x-1)+log_{4}(2x-5) is log_{4}((x-1)(2x-5))

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simplify log_{4}(x-1)+log_{4}(2x-5)

simplify

lo g_{4} (x - 1) + lo g_{4} (2 x - 5)

lo g_{4} ((x - 1) (2 x - 5))

Solution steps

lo g_{4} (x - 1) + lo g_{4} (2 x - 5)

Apply log rule:

lo g_{c} (a) + lo g_{c} (b) = lo g_{c} (ab)

lo g_{4} (x - 1) + lo g_{4} (2 x - 5) = lo g_{4} ((x - 1) (2 x - 5))

= lo g_{4} ((x - 1) (2 x - 5))

ln (\frac{2 e}{3})

- lo g_{10} (5.2 \cdot 1 0^{- 4})

\frac{ln ( 37 ) + 2 ln ( 4 )}{ln ( 4 )}

ln (\frac{e ^{2 x} cos ( 3 x )}{x - 4})

\frac{π}{8} ln (\frac{13}{9})

What is log_{4}(x-1)+log_{4}(2x-5) ?
The solution to log_{4}(x-1)+log_{4}(2x-5) is log_{4}((x-1)(2x-5))