What is the answer to a=log_{3210}(8)+log_{23}(x) ?

The answer to a=log_{3210}(8)+log_{23}(x) is x=23^{a-3log_{3210}(2)}

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a=log_{3210}(8)+log_{23}(x)

a = lo g_{3210} (8) + lo g_{23} (x)

x = 2 3^{a - 3 l o g_{3210} (2)}

Solution steps

a = lo g_{3210} (8) + lo g_{23} (x)

Switch sides

lo g_{3210} (8) + lo g_{23} (x) = a

Move

lo g_{3210} (8)

to the right side

lo g_{23} (x) = a - lo g_{3210} (8)

Apply log rules

x = 2 3^{a - 3 l o g_{3210} (2)}

Verify Solutions:

x = 2 3^{a - 3 l o g_{3210} (2)}

True

The solution is

x = 2 3^{a - 3 l o g_{3210} (2)}

lo g_{3} (2 x - 11) = 2

lo g_{6} (3 x) - lo g_{6} (4 x - 1) = 1

lo g_{10} (2 x + 3) = lo g_{10} (7)

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

x^{2 \cdot l o g_{10} (x)} = 10 x^{2}

What is the answer to a=log_{3210}(8)+log_{23}(x) ?
The answer to a=log_{3210}(8)+log_{23}(x) is x=23^{a-3log_{3210}(2)}