What is the answer to 2log_{10}(4x)-log_{10}(49)=2 ?

The answer to 2log_{10}(4x)-log_{10}(49)=2 is x=(5^{(2+2log_{10}(7))/2})/(2^{(2-2log_{10)(7))/2}}

Solve for x: 2log_{10}(4x)-log_{10}(49)=2

The solution is x=(5^{(2+2log_{10}(7))/2})/(2^{(2-2log_{10)(7))/2}}

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2 lo g_{10} (4 x) - lo g_{10} (49) = 2

x = \frac{5 ^{\frac{2 + 2 l o g _{10} ( 7 )}{2}}}{2 ^{\frac{2 - 2 l o g _{10} ( 7 )}{2}}}

Decimal Notation

x = 17.5

Solution steps

2 lo g_{10} (4 x) - lo g_{10} (49) = 2

Move

lo g_{10} (49)

to the right side

2 lo g_{10} (4 x) = 2 + lo g_{10} (49)

Divide both sides by

2

lo g_{10} (4 x) = \frac{2 + lo g _{10} ( 49 )}{2}

Apply log rules

4 x = 1 0^{\frac{2 + 2 l o g _{10} ( 7 )}{2}}

Solve

4 x = 1 0^{\frac{2 + 2 l o g _{10} ( 7 )}{2}} : x = \frac{5 ^{\frac{2 + 2 l o g _{10} ( 7 )}{2}}}{2 ^{\frac{2 - 2 l o g _{10} ( 7 )}{2}}}

x = \frac{5 ^{\frac{2 + 2 l o g _{10} ( 7 )}{2}}}{2 ^{\frac{2 - 2 l o g _{10} ( 7 )}{2}}}

Verify Solutions:

x = \frac{5 ^{\frac{2 + 2 l o g _{10} ( 7 )}{2}}}{2 ^{\frac{2 - 2 l o g _{10} ( 7 )}{2}}}

True

The solution is

x = \frac{5 ^{\frac{2 + 2 l o g _{10} ( 7 )}{2}}}{2 ^{\frac{2 - 2 l o g _{10} ( 7 )}{2}}}

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

\frac{3}{2} = lo g_{4} (x)

lo g_{10} (25 x) = 3

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

ln (3 x - 12) = 7

What is the answer to 2log_{10}(4x)-log_{10}(49)=2 ?
The answer to 2log_{10}(4x)-log_{10}(49)=2 is x=(5^{(2+2log_{10}(7))/2})/(2^{(2-2log_{10)(7))/2}}
Solve for x: 2log_{10}(4x)-log_{10}(49)=2
The solution is x=(5^{(2+2log_{10}(7))/2})/(2^{(2-2log_{10)(7))/2}}