If [tex]a = log_92[/tex] and [tex]b = log_54[/tex], find [tex]log_615[/tex] in terms of a and b.
Please answer with solution, thanks in advance! :))
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If [tex]a = log_92[/tex] and [tex]b = log_54[/tex], find [tex]log_615[/tex] in terms of a and b.
Please answer with solution, thanks in advance! :))
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Answer:
[tex]\sf log_615=\dfrac{log(15)}{log(6)}[/tex]
[tex]\sf log_615=\dfrac{log(5 \times 3)}{log(2 \times 3)}[/tex]
[tex]\sf log_615=\dfrac{log5 + log3}{log2 + log3}[/tex]
Now,
[tex]\sf log_92=a=\dfrac{log2}{log9}[/tex]
[tex]\sf log2=a(log9)[/tex]
Also,
[tex]\sf log_54=b=\dfrac{log4}{log5}[/tex]
[tex]\sf log5=\dfrac{log_4}{b}[/tex]
So, substitute these in the equation and lets see what happens.
[tex]\sf log_615=\dfrac{\dfrac{log2^2}{b} + log3}{a(log3^2) + log3}[/tex]
[tex]\sf log_615=\dfrac{\dfrac{2log2}{b} + log3}{a(2log3) + log3}[/tex]
[tex]\sf log_615=\dfrac{\dfrac{2log2}{b} + log3}{log3(2a + 1)}[/tex]
[tex]\sf log_615=\dfrac{log3(\dfrac{2log2}{blog3} + 1)}{log3(2a + 1)}[/tex]
[tex]\sf log_615=\dfrac{(\dfrac{2log2}{blog3} + 1)}{(2a + 1)}[/tex]
[tex]\sf log_615=\dfrac{ (\dfrac{2}{b} \cdot \dfrac{log2}{log3} + 1)}{(2a + 1)}[/tex]
a = 1/2(log₃2)
2a = log₃2
[tex]\sf log_615=\dfrac{ (\dfrac{2}{b} \cdot 2a + 1)}{(2a + 1)}[/tex]
[tex]\sf log_615=\dfrac{ (\dfrac{4a}{b} + 1)}{(2a + 1)}[/tex]
#CarryOnLearning :D
Given :
To Find :
Solution :
[tex] \sf{log_92 = a}[/tex]
[tex]\implies\sf\dfrac{log2}{log9} = a[/tex]
[tex]\implies\sf{\dfrac{1}{2} \times \dfrac{log2}{log3}} = a[/tex]
[tex]\implies\sf log3 = \dfrac{1}{2} \times \dfrac{log2}{a}[/tex]
[tex] \\ [/tex]
[tex]\sf{log_54 = b}[/tex]
[tex]\implies\sf\dfrac{log4}{log5} = b[/tex]
[tex]\implies\sf log5 = 2 \times \dfrac{log2}{b}[/tex]
Now,
[tex]\sf{log_615}[/tex]
[tex]= \sf\dfrac{log15}{log6}[/tex]
[tex]= \sf\dfrac{log5 + log3}{log2 + log3}[/tex]
[tex] \\ [/tex]
We have,
Substituting as follows,
[tex]= \sf{\dfrac{\dfrac{2 \times log2}{b} + \dfrac{1}{2} \times \dfrac{log2}{a}}{log2 + \dfrac{1}{2} \times \dfrac{log2}{a}}[/tex]
[tex]= \sf{\dfrac{\sf{\dfrac{2}{b} + \dfrac{1}{2a}}}{1 + \dfrac{1}{2a}}[/tex]
[tex]= \sf{\dfrac{4a + b}{2ab} \div \dfrac{2a + 1}{2a}}[/tex]
[tex]= \sf{\dfrac{4a + b}{2ab} \times \dfrac{2a}{2a + 1}}[/tex]
[tex]= \sf{\dfrac{4a +b}{2ab + b}}[/tex]
[tex] \\ [/tex]
Request :
☆ Kindly see the answer from web
#CarryOnLearning