The logarithmic expression

a1" title="log_{b}(\frac{1}{b^{-b}} )" alt="log_{b}(\frac{1}{b^{-b}} )" align="absmiddle" class="latex-formula"> where b > 0, is equal to
a) $b^{-b}$

b) $b^{b}$

c) $-b$

d) $b$

Mathematics

1 answer:

ivanzaharov [21]1 year ago

3 0

$\log_b \left(\dfrac 1{b^{-b}}\right)\\\\=\log_b (b^{-b})^{-1}\\\\=\log_b b^b\\\\=b \log_b b\\\\=b \cdot 1\\\\=b$

You might be interested in

If C = 1+ p2 + 3p and B = 2p + 6p, find an expression that equals 2C – – B in

drek231 [11]

Answer:

2p² + 14p + 2

Step-by-step explanation:

Given that;

C= 1+p²+3p

B=2p+6p

The expression that equals to : 2C - -B will be;

2{1+p²+3p} - {-2p-6p}

2+2p²+6p + 2p + 6p

2p² + 6p +2p +6p +2

2p² + 14p + 2

4 0

2 years ago

Jason eats 10 ounces of candy in 5 days. a. How many pounds does he eat per day?

Nadusha1986 [10]

A. He eats 1/8 a pound of candy each day.
B. It would take him a total of 8 days to eat a pound of candy.

A. Since he ate 10 ounces in 5 days, you divide 10 by 5 to find he ate 2 ounces per day. Since there are 16 ounces in a pound that would be your denominator (bottom of fraction) and 2 would be the numerator (top of your fraction). You will get 2/16 which if you simplify (divide by 2) you get 1/8.

B. Knowing that there are 16 ounces per pound and he ate 1/8 pound ( 2/16 pound or 2 ounces) a day, you can either look at the simplified denominator to get the answer, or, divide 16 by 2 to get your answer.

I hope this helps!

8 0

2 years ago

Flying to Kimpala with a tail wind a plane averaged 158km/hr. On the return trip the plane only average 112km/hr. While flying b

lora16 [44]

The speed of wind and speed of plane in still air are 23 and 135 km/h respectively.

<u>Step-by-step explanation:</u>

Let the speed of wind and speed of plane in still air are w and p km/h respectively.

The effective speed on onward journey was

$p+w=158$ ................(1)