What is log_{3}(y)+y<4 ?

The solution to log_{3}(y)+y<4 is 0<y<3

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log_{3}(y)+y<4

lo g_{3} (y) + y < 4

0 < y < 3

Interval Notation

(0, 3)

Solution steps

lo g_{3} (y) + y < 4

Move

y

to the right side

lo g_{3} (y) < 4 - y

For a>1, if

f (x) < g (x),

then

a^{f (x)} < a^{g (x)}

a = 3

3^{l o g_{3} (y)} < 3^{4 - y}

Simplify

3^{l o g_{3} (y)} : y

y < 3^{4 - y}

Solve

y < 3^{4 - y} : y < 3

Find positive values for logs:

y > 0

Combine the intervals

y < 3 and y > 0

Merge Overlapping Intervals

0 < y < 3

ln (x - 1) \geq - 1

ln (e^{x} - 1) > 0

lo g_{3} (lo g_{10} (3 x)) < lo g_{3} (81)

lo g_{2} (x^{2} + 3 x + 2) \leq 1

lo g_{2} (2 + 3 x) \leq 4