What is the answer to log_{a}(5y-4)-log_{a}(2y)=3 ?

The answer to log_{a}(5y-4)-log_{a}(2y)=3 is y= 4/(5-2a^3) {0<a<1\lor 1<a<\sqrt[3]{5/2}}

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log_{a}(5y-4)-log_{a}(2y)=3

lo g_{a} (5 y - 4) - lo g_{a} (2 y) = 3

Solution steps

lo g_{a} (5 y - 4) - lo g_{a} (2 y) = 3

Add

lo g_{a} (2 y)

to both sides

lo g_{a} (5 y - 4) - lo g_{a} (2 y) + lo g_{a} (2 y) = 3 + lo g_{a} (2 y)

Simplify

lo g_{a} (5 y - 4) = 3 + lo g_{a} (2 y)

Apply log rules

5 y - 4 = a^{3} \cdot 2 y

Solve

5 y - 4 = a^{3} \cdot 2 y : y = \frac{4}{5 - 2 a ^{3}}

y = \frac{4}{5 - 2 a ^{3}}

Verify Solutions:

The solution is

- 4 lo g_{3} (3 x + 8) - 2 = - 2

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

lo g_{x} (6) = 7

0 = 1.88 + (- 0.225) ln (x)

ln (3 x) - 2 = 5

What is the answer to log_{a}(5y-4)-log_{a}(2y)=3 ?
The answer to log_{a}(5y-4)-log_{a}(2y)=3 is y= 4/(5-2a^3) {0<a<1\lor 1<a<\sqrt[3]{5/2}}