What is 0.8^{log_{125}(x)}>= 0.8^{log_{10}(125)} ?

The solution to 0.8^{log_{125}(x)}>= 0.8^{log_{10}(125)} is 0<x<= 1953125^{log_{10}(5)}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

0.8^{log_{125}(x)}>= 0.8^{log_{10}(125)}

0. 8^{l o g_{125} (x)} \geq 0. 8^{l o g_{10} (125)}

0 < x \leq 195312 5^{l o g_{10} (5)}

Interval Notation

(0, 195312 5^{l o g_{10} (5)}]

Decimal Notation

0 < x \leq 24947.76259 \dots

Solution steps

0. 8^{l o g_{125} (x)} \geq 0. 8^{l o g_{10} (125)}

Let

u = lo g_{125} (x)

0. 8^{u} \geq 0. 8^{l o g_{10} (125)}

0. 8^{u} \geq 0. 8^{l o g_{10} (125)} : u \leq 3 lo g_{10} (5)

u \leq 3 lo g_{10} (5)

Substitute back

u = lo g_{125} (x)

lo g_{125} (x) \leq 3 lo g_{10} (5)

Apply log rules

x \leq 195312 5^{l o g_{10} (5)} and x > 0

Merge Overlapping Intervals

0 < x \leq 195312 5^{l o g_{10} (5)}

ln (x - e) > 0

(lo g_{2} (\frac{( x - 1 )}{( x + 2 )})) > 0

lo g_{3} (2 x + 5) < 2

(3^{x^{2} - 1} - 2 7^{x + 1}) (lo g_{3} (x + 8) - 2) \leq 0

lo g_{3} (x^{2}) > 2

What is 0.8^{log_{125}(x)}>= 0.8^{log_{10}(125)} ?
The solution to 0.8^{log_{125}(x)}>= 0.8^{log_{10}(125)} is 0<x<= 1953125^{log_{10}(5)}