What is the answer to log_{10}(6x)+ln(5x)=0 ?

The answer to log_{10}(6x)+ln(5x)=0 is x=10^{(-log_{10}(6)log_{10}(e)-log_{10}(5))/(log_{10)(e)+1}}

Solve for x: log_{10}(6x)+ln(5x)=0

The solution is x=10^{(-log_{10}(6)log_{10}(e)-log_{10}(5))/(log_{10)(e)+1}}

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lo g_{10} (6 x) + ln (5 x) = 0

x = 1 0^{\frac{- l o g _{10} ( 6 ) l o g _{10} ( e ) - l o g _{10} ( 5 )}{l o g _{10} ( e ) + 1}}

Decimal Notation

x = 0.18925 \dots

Solution steps

lo g_{10} (6 x) + ln (5 x) = 0

Apply log rules

x = 1 0^{\frac{- l o g _{10} ( 6 ) l o g _{10} ( e ) - l o g _{10} ( 5 )}{l o g _{10} ( e ) + 1}}

Verify Solutions:

x = 1 0^{\frac{- l o g _{10} ( 6 ) l o g _{10} ( e ) - l o g _{10} ( 5 )}{l o g _{10} ( e ) + 1}}

True

The solution is

x = 1 0^{\frac{- l o g _{10} ( 6 ) l o g _{10} ( e ) - l o g _{10} ( 5 )}{l o g _{10} ( e ) + 1}}

lo g_{8} (x) + lo g_{8} (x + 12) = 2

lo g_{2} (x + 4) = 1 + lo g_{2} (x - 3)

ln (3 - x^{2}) = 2 ln (x)

lo g_{2} (4 x) - lo g_{2} (x^{2}) = 4

2 x \cdot ln (x) + x = 0

What is the answer to log_{10}(6x)+ln(5x)=0 ?
The answer to log_{10}(6x)+ln(5x)=0 is x=10^{(-log_{10}(6)log_{10}(e)-log_{10}(5))/(log_{10)(e)+1}}
Solve for x: log_{10}(6x)+ln(5x)=0
The solution is x=10^{(-log_{10}(6)log_{10}(e)-log_{10}(5))/(log_{10)(e)+1}}