What is the answer to 282=122log_{10}(0.8x)+85 ?

The answer to 282=122log_{10}(0.8x)+85 is x=(25*5^{75/122})/(2^{47/122)}

Solve for x: 282=122log_{10}(0.8x)+85

The solution is x=(25*5^{75/122})/(2^{47/122)}

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282 = 122 lo g_{10} (0.8 x) + 85

x = \frac{25 \cdot 5 ^{\frac{75}{122}}}{2 ^{\frac{47}{122}}}

Decimal

x = 51.48303 \dots

Solution steps

282 = 122 lo g_{10} (0.8 x) + 85

Switch sides

122 lo g_{10} (0.8 x) + 85 = 282

Move

85

to the right side

122 lo g_{10} (0.8 x) = 197

Divide both sides by

122

lo g_{10} (0.8 x) = \frac{197}{122}

Apply log rules

0.8 x = 10 \cdot 1 0^{\frac{75}{122}}

Solve

0.8 x = 10 \cdot 1 0^{\frac{75}{122}} : x = \frac{25 \cdot 5 ^{\frac{75}{122}}}{2 ^{\frac{47}{122}}}

x = \frac{25 \cdot 5 ^{\frac{75}{122}}}{2 ^{\frac{47}{122}}}

Verify Solutions:

x = \frac{25 \cdot 5 ^{\frac{75}{122}}}{2 ^{\frac{47}{122}}}

True

The solution is

x = \frac{25 \cdot 5 ^{\frac{75}{122}}}{2 ^{\frac{47}{122}}}

ln (x) = 5.2339583

lo g_{3} (5 x + 1) = 4

lo g_{2} (x) = 5

lo g_{2} (x - 8) = 4

lo g_{10} (x^{2}) = lo g_{10} (10) + lo g_{10} (x) - lo g_{10} (9) + lo g_{10} (10)

What is the answer to 282=122log_{10}(0.8x)+85 ?
The answer to 282=122log_{10}(0.8x)+85 is x=(25*5^{75/122})/(2^{47/122)}
Solve for x: 282=122log_{10}(0.8x)+85
The solution is x=(25*5^{75/122})/(2^{47/122)}