What is the answer to log_{4}(2x)+3log_{4}(7)=3 ?

The answer to log_{4}(2x)+3log_{4}(7)=3 is x= 32/343

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log_{4}(2x)+3log_{4}(7)=3

lo g_{4} (2 x) + 3 lo g_{4} (7) = 3

x = \frac{32}{343}

Decimal Notation

x = 0.09329 \dots

Solution steps

lo g_{4} (2 x) + 3 lo g_{4} (7) = 3

Move

3 lo g_{4} (7)

to the right side

lo g_{4} (2 x) = 3 - 3 lo g_{4} (7)

Apply log rules

2 x = \frac{64}{343}

Solve

2 x = \frac{64}{343} : x = \frac{32}{343}

x = \frac{32}{343}

Verify Solutions:

x = \frac{32}{343}

True

The solution is

x = \frac{32}{343}

\frac{2}{x ln ( x )} = 0

lo g_{10} (x) = 2.2

v, lo g_{q} (v) = b

lo g_{3} (7 - 4 x) - lo g_{3} (x - 5) = 2

3 x lo g_{10} (x) = 9

What is the answer to log_{4}(2x)+3log_{4}(7)=3 ?
The answer to log_{4}(2x)+3log_{4}(7)=3 is x= 32/343
Solve for x: log_{4}(2x)+3log_{4}(7)=3
The solution is x= 32/343