The given inequalities can be evaluated using the rules of indices and
logarithms.
Correct responses:
- 2·(5)ˣ > 250 [tex]\longrightarrow[/tex] x > 3
- [tex]4 \cdot \left(\frac{1}{3} \right)^x < 36 \longrightarrow \underline{ x < -2}[/tex]
- [tex]10 \cdot \left(\frac{1}{2} \right)^x > 80 \longrightarrow \underline{x > -3 }[/tex]
- [tex]\left(\frac{1}{2} \right) \cdot (6)^x < 18 \longrightarrow \underline{ x < 2}[/tex]
Methods used for the evaluation of the inequalities
(1) 2(5)ˣ > 250, gives;
[tex]5^x > \mathbf{\dfrac{250}{2}} = 125[/tex]
x·㏑5 > ㏑125
[tex]x > \dfrac{ln(125)}{ln(5)} = \dfrac{3 \cdot ln(5)}{ln(5)} = \mathbf{3}[/tex]
x > 3
- 2·(5)ˣ > 250 [tex]\longrightarrow[/tex] x > 3
(2) [tex]\mathbf{4 \cdot \left(\frac{1}{3} \right)^x } < 36[/tex]
[tex]\left(\frac{1}{3} \right)^x < \dfrac{36}{4} = 9[/tex]
[tex]\mathbf{x \cdot ln \left(\frac{1}{3} \right)} < ln(9)[/tex]
[tex]x < \dfrac{ln(9)}{ln\left(\frac{1}{3} \right)} = \dfrac{2 \cdot ln(3)}{-ln(3)} = \mathbf{-2}[/tex]
x < -2
- [tex]4 \cdot \left(\frac{1}{3} \right)^x < 36 \longrightarrow \underline{ x < -2}[/tex]
(3) [tex]\mathbf{10 \cdot \left(\frac{1}{2} \right)^x } > 80[/tex]
[tex]\left(\frac{1}{2} \right)^x > \mathbf{ \dfrac{80}{10} }= 8[/tex]
[tex]x \cdot ln \left(\frac{1}{2} \right) >ln(8)[/tex]
[tex]x > \mathbf{\dfrac{ln(8)}{ln\left(\frac{1}{2} \right)} }= \dfrac{3 \cdot ln(2)}{-1 \cdot ln(2)} = -3[/tex]
x > -3
- [tex]10 \cdot \left(\frac{1}{2} \right)^x > 80 \longrightarrow \underline{x > -3}[/tex]
(4) [tex]\mathbf{\left(\frac{1}{2} \right) \cdot (6)^x} < 18[/tex]
[tex]6^x < \dfrac{18}{\dfrac{1}{2} } = \mathbf{36}[/tex]
6ˣ < 36 = 6²
6ˣ < 6²
x < 2
- [tex]\left(\frac{1}{2} \right) \cdot (6)^x < 18 \longrightarrow \underline{x < 2}[/tex]
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