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  • Inequation:
  • a*x>3
  • abs(2*x*x)<1
  • x-2/(x-5)(x-1)≥0
  • (x^2-26)^2=<100
  • Sum of series:
  • 17 17
  • Derivative of:
  • 17 17
  • Limit of the function:
  • 17 17

  • Identical expressions

  • eight ^x+ two ^(four - three *x)> seventeen
  • 8 to the power of x plus 2 to the power of (4 minus 3 multiply by x) greater than 17
  • eight to the power of x plus two to the power of (four minus three multiply by x) greater than seventeen
  • 8x+2(4-3*x)>17
  • 8x+24-3*x>17
  • 8^x+2^(4-3x)>17
  • 8x+2(4-3x)>17
  • 8x+24-3x>17
  • 8^x+2^4-3x>17

  • Similar expressions

  • 8^x-2^(4-3*x)>17
  • 8^x+2^(4+3*x)>17
Solving inequalities/ 8^x+2^(4-3*x)>17

8^x+2^(4-3*x)>17 inequation

A inequation with variable

The solution

You have entered [src] 
 x    4 - 3*x     
8  + 2        > 17
$$2^{4 - 3 x} + 8^{x} > 17$$ 
2^(4 - 3*x) + 8^x > 17
Detail solution
Given the inequality:
$$2^{4 - 3 x} + 8^{x} > 17$$
To solve this inequality, we must first solve the corresponding equation:
$$2^{4 - 3 x} + 8^{x} = 17$$
Solve:
Given the equation:
$$2^{4 - 3 x} + 8^{x} = 17$$
or
$$\left(2^{4 - 3 x} + 8^{x}\right) - 17 = 0$$
Do replacement
$$v = \left(\frac{1}{8}\right)^{x}$$
we get
$$2^{4 - 3 x} - 17 + \frac{1}{v} = 0$$
or
$$2^{4 - 3 x} - 17 + \frac{1}{v} = 0$$
do backward replacement
$$\left(\frac{1}{8}\right)^{x} = v$$
or
$$x = - \frac{\log{\left(v \right)}}{\log{\left(8 \right)}}$$
$$x_{1} = 0$$
$$x_{2} = \frac{4}{3}$$
$$x_{3} = - \frac{2 i \pi}{3 \log{\left(2 \right)}}$$
$$x_{4} = \frac{2 i \pi}{3 \log{\left(2 \right)}}$$
$$x_{5} = \frac{4}{3} - \frac{2 i \pi}{3 \log{\left(2 \right)}}$$
$$x_{6} = \frac{4}{3} + \frac{2 i \pi}{3 \log{\left(2 \right)}}$$
Exclude the complex solutions:
$$x_{1} = 0$$
$$x_{2} = \frac{4}{3}$$
This roots
$$x_{1} = 0$$
$$x_{2} = \frac{4}{3}$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} < x_{1}$$
For example, let's take the point
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10}$$
=
$$- \frac{1}{10}$$
substitute to the expression
$$2^{4 - 3 x} + 8^{x} > 17$$
$$\frac{1}{\sqrt[10]{8}} + 2^{4 - \frac{\left(-1\right) 3}{10}} > 17$$
 7/10                
2           3/10     
----- + 16*2     > 17
  2

one of the solutions of our inequality is:
$$x < 0$$
 _____           _____          
      \         /
-------ο-------ο-------
       x1      x2

Other solutions will get with the changeover to the next point
etc.
The answer:
$$x < 0$$
$$x > \frac{4}{3}$$
Solving inequality on a graph
Rapid solution 2 [src] 
               /  3 ___\     
            log\2*\/ 2 /     
(-oo, 0) U (------------, oo)
               log(2)
$$x\ in\ \left(-\infty, 0\right) \cup \left(\frac{\log{\left(2 \sqrt[3]{2} \right)}}{\log{\left(2 \right)}}, \infty\right)$$ 
x in Union(Interval.open(-oo, 0), Interval.open(log(2*2^(1/3))/log(2), oo))
Rapid solution [src] 
  /          /  3 ___\    \
  |       log\2*\/ 2 /    |
Or|x < 0, ------------ < x|
  \          log(2)       /
$$x < 0 \vee \frac{\log{\left(2 \sqrt[3]{2} \right)}}{\log{\left(2 \right)}} < x$$ 
(x < 0)∨(log(2*2^(1/3))/log(2) < x)
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