3 years ago

7 log(15/16)+6log(8/3)+5log(2/3)+log(32/25)by logarithm laws

Mathematics

1 answer:

hammer [34]3 years ago

4 0

Answer:

1.5871

Step-by-step explanation:

According to the Question,

we have, 7 log(15/16)+6log(8/3)+5log(2/3)+log(32/25)

The Solution, With Basic Properties of Log.

7*log15-7*log16+6*log8-6*log3+5*log2-5*log3+log32-log25

7*1.176-7*1.204+6*0.903-6*0.477+5*0.3010-5*0.477+1.505-1.3979

on Solving, We get

1.5871

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If (x)=3x-1 and s(x)=2x+1, which expression is equivalent to r over s (6) ?

katrin [286]

Answer:

$\large\boxed{\dfrac{3(6)-1}{2(6)+1}}$

Step-by-step explanation:

$\left(\dfrac{f}{g}\right)(x)=\dfrac{f(x)}{g(x)}\\\\\text{We have}\\\\f(x)=3x-1,\ g(x)=2x+1\\\\\left(\dfrac{f}{g}\right)(x)=\dfrac{3x-1}{2x+1}\\\\\left(\dfrac{f}{g}\right)(6)\to\text{put x = 6 to the equation of the function}\ \left(\dfrac{f}{g}\right)(x):\\\\\left(\dfrac{f}{g}\right)(6)=\dfrac{3(6)-1}{2(6)+1}$