What is the answer to log_{y}(2)-log_{4}(y)=1 ?

The answer to log_{y}(2)-log_{4}(y)=1 is y= 1/(2^{1+sqrt(3))},y=2^{sqrt(3)-1}

Solve for y: log_{y}(2)-log_{4}(y)=1

The solution is y= 1/(2^{1+sqrt(3))},y=2^{sqrt(3)-1}

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lo g_{y} (2) - lo g_{4} (y) = 1

y = \frac{1}{2 ^{1 + 3}}, y = 2^{3 - 1}

Decimal Notation

y = 0.15051 \dots, y = 1.66099 \dots

Solution steps

lo g_{y} (2) - lo g_{4} (y) = 1

Apply log rules

\frac{1}{2 lo g _{4} ( y )} - lo g_{4} (y) = 1

Rewrite the equation with

lo g_{4} (y) = u

\frac{1}{2 u} - u = 1

Solve

\frac{1}{2 u} - u = 1 : u = - \frac{1 + 3}{2}, u = \frac{3 - 1}{2}

u = - \frac{1 + 3}{2}, u = \frac{3 - 1}{2}

Substitute back

u = lo g_{4} (y),

solve for

y

Solve

lo g_{4} (y) = - \frac{1 + 3}{2} : y = \frac{1}{2 ^{1 + 3}}

Solve

lo g_{4} (y) = \frac{3 - 1}{2} : y = 2^{3 - 1}

y = \frac{1}{2 ^{1 + 3}}, y = 2^{3 - 1}

Verify Solutions:

y = \frac{1}{2 ^{1 + 3}}

True

, y = 2^{3 - 1}

True

The solutions are

y = \frac{1}{2 ^{1 + 3}}, y = 2^{3 - 1}

lo g_{4} (1 + x^{2}) = - 0.25

lo g_{10} (y) = - 138.265

lo g_{2} (\frac{1}{x}) = - 1

ln (x - 2) - ln (8) = 2

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

What is the answer to log_{y}(2)-log_{4}(y)=1 ?
The answer to log_{y}(2)-log_{4}(y)=1 is y= 1/(2^{1+sqrt(3))},y=2^{sqrt(3)-1}
Solve for y: log_{y}(2)-log_{4}(y)=1
The solution is y= 1/(2^{1+sqrt(3))},y=2^{sqrt(3)-1}