"log⁡15(5)=alog⁡15(2)=b Note: 15 is the log base in..." solved and explained

sodni3

Answered question

2021-09-25

\log 15 (5) = a \log 15 (2) = b

Note: 15 is the log base in this case Express

\log 15 (360)

in terms of a and b. The final answer cannot have logs, fractions, or decimals.

Jozlyn

Skilled2021-09-26Added 85 answers

Step 1
As

\log_{15} (5) = a \dots (1)

\log_{15} (2) = b \dots (2)

From (1)

a = \log_{15} (5)

a = \frac{1}{\log_{5} 15} [∵ \log_{a} b = \log_{b} a]

a = \frac{1}{\log_{5} (5 \times 3)} = \frac{1}{\log_{5} 5 + \log_{5} 3}

a = \frac{1}{1 + \log_{5} 3} [∵ \log_{a} a = 1]

1 + \log_{5} 3 = \frac{1}{a}

\log_{5} 3 = \frac{1}{a} - 1

\log_{5} 3 = \frac{1 - a}{a}

\log_{3} 5 = \frac{a}{1 - a} \dots (3)

Step 2
Now,

\log_{15} 360 = \log_{15} (2^{3} \times 5 \times 3^{2})

= \log_{15} 2^{3} + \log_{15} 5 + \log_{15} 3^{2}

= 3 \log_{15}^{2} + \log_{15} 5 + 2 \log_{15} 3

= 3 b + a + \frac{2}{\log_{3} 15}

= 3 b + a + \frac{2}{\log_{3} (3 \times 5)}

= 3 b + a + \frac{2}{\log_{3} 3 + \log_{3} 5}

= 3 b + a + \frac{2}{1 + \log_{3} 5}

= 3 b + a + \frac{2}{1 + \frac{a}{1 - a}}

(from(3))

= 3 b + a + \frac{2}{(\frac{1 - a + a}{1 - a})}

= 3 b + a + 2 (1 - a)

= 3 b +

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