What is 2^{log_{2}(x+1)}<8 ?

The solution to 2^{log_{2}(x+1)}<8 is -1<x<7

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2^{log_{2}(x+1)}<8

2^{l o g_{2} (x + 1)} < 8

- 1 < x < 7

Interval Notation

(- 1, 7)

Solution steps

2^{l o g_{2} (x + 1)} < 8

Apply log rules

x + 1 < 8

Move

1

to the right side

x < 7

Find positive values for logs:

x > - 1

Combine the intervals

x < 7 and x > - 1

Merge Overlapping Intervals

- 1 < x < 7

x^{4} + 4 < 3 or - x + 2 < 2

2 x = x - 6

\frac{4 !}{( 4 - 2 ) !}

\frac{x}{4} - \frac{1}{20} \leq \frac{x}{5} + 1

7 - (- 2)