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Sagot :
You mean situation, when a^b < a.
There are two ways:
[tex]a^b < a^1 \\ b<1 [/tex]
But exponent function a^b has to be increasing so a>1
So first solution: a>1 and 0<b<1. Examples:
[tex]5^{\frac{1}{2}} < 5 \\ 120^{\frac{1}{12}} < 120[/tex]
and second way:
[tex]a^b < a^1 \\ b>1[/tex]
Exponent function a^b has to be decreasing, so 0<a<1
Second solution: a<1 and b>1. Examples:
[tex]\left(\dfrac{1}{2}\right)^{12} < \dfrac{1}{2} \\ \\ \left(\dfrac{12}{39}\right)^{16\frac{1}{2}} < \dfrac{12}{39}[/tex]
There are two ways:
[tex]a^b < a^1 \\ b<1 [/tex]
But exponent function a^b has to be increasing so a>1
So first solution: a>1 and 0<b<1. Examples:
[tex]5^{\frac{1}{2}} < 5 \\ 120^{\frac{1}{12}} < 120[/tex]
and second way:
[tex]a^b < a^1 \\ b>1[/tex]
Exponent function a^b has to be decreasing, so 0<a<1
Second solution: a<1 and b>1. Examples:
[tex]\left(\dfrac{1}{2}\right)^{12} < \dfrac{1}{2} \\ \\ \left(\dfrac{12}{39}\right)^{16\frac{1}{2}} < \dfrac{12}{39}[/tex]
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