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3 years ago

Evaluate the following expression. Round your answer two decimal places. Log3(10)

Mathematics

2 answers:

Alexandra [31]3 years ago

6 0

on apex it says that its B: 2.10

Elanso [62]3 years ago

3 0

$\bf \textit{Logarithm Change of Base Rule}
\\\\
log_a b\implies \cfrac{log_c b}{log_c a}\\\\
-------------------------------\\\\
log_3(10)\implies \cfrac{log_{10}(10)}{log_{10}(3)}\implies \cfrac{1}{log(3)}$

and you can just plug that in your calculator.

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MA_775_DIABLO [31]

Answer:

1/10

when you do the LCM you get 70

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2 years ago

Which of the following expressions are equivalent to this expression:

Gekata [30.6K]

The original expression is equal to 0 because anything multiplied by 0 is equal to 0. Solve inside the brackets for the possible answer choices to find what will equal 0.

Start with the first expression. Add 4 and negative 4 will become 0, and -1 times 0 is equal to 0. Let's solve for the others just to be sure.

In the second expression, solving inside the brackets gives you 8. -1 times 8 is equal to -8.

Adding 4 and negative 4 in the third expression leaves you with 0. But, 1 + 0 is equal to 1.

Adding negative 4 and negative 4 gives you the answer of -8, and -1 times -8 is equal to 8.

Your answer is the first expression, or A.

6 0

3 years ago

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How to solve -35g = -8h + 11k (solving for h)?

Fed [463]

Answer:

$h = \frac{35}{g} +\frac{11k}{8}$

Step-by-step explanation:

8 0

2 years ago

Find the limit. Use l'Hospital's Rule if appropriate. If there is a more elementary method, consider using it. lim x→1 3x x − 1

Reika [66]

Answer:

Step-by-step explanation:

Given the limit of a function expressed as $\lim_{x \to 1} (\dfrac{3x}{x-1} - \dfrac{3}{lnx})$ , we are to evaluate it. To evaluate it, we will simply substitute x = 1 into the function since the variable x tends to 1.

$\lim_{x \to 1} (\dfrac{3x}{x-1} - \dfrac{3}{lnx})\\\\= (\dfrac{3(1)}{1-1} - \dfrac{3}{ln1})\\\\= \dfrac{3}{0} - \dfrac{3}{0}\\\\= \infty - \infty (ind)$

Since we got an indeterminate function, we will find the LCM of the function and solve again.

$= \lim_{x \to 1} (\dfrac{3x}{x-1} - \dfrac{3}{lnx})\\\\= \lim_{x \to 1} \dfrac{3xlnx-3(x-1)}{(x-1)lnx}\\\\\\= \dfrac{3(1)ln(1)-3(1-1)}{(1-1)ln1}\\\\= \frac{3(0)-3(0)}{0(0)} \\\\= \frac{0}{0} (ind)$

Applying L'hospital rule;

$\frac{x}{y} = \lim_{x \to 1} \dfrac{d/dx(3xlnx-3(x-1))}{d/dx((x-1)lnx)}\\\\= \lim_{x \to 1} \dfrac{3x(\frac{1}{x})+ 3lnx-3)}{(x-1)\frac{1}{x} +lnx}\\\\= \lim_{x \to 1} \dfrac{3 + 3lnx-3}{(x-1)\frac{1}{x} +lnx}\\\\= \frac{3ln1}{(1-1)\frac{1}{1} +ln1}\\\\= \frac{0}{0} (ind)$

Applying L'hospital rule again;

$= \lim_{x \to 1} \dfrac{\frac{d}{dx} (3lnx)}{\frac{d}{dx} ((x-1)\frac{1}{x} +lnx)}\\\\= \lim_{x \to 1} \dfrac{\frac{3}{x} }{(x-1)\frac{-1}{x^2} + \frac{1}{x} +\frac{1}{x} }\\\\= \dfrac{\frac{3}{1} }{(1-1)\frac{-1}{1^2} + \frac{1}{1} +\frac{1}{1} }\\\\= \frac{3}{0(-1)+2}\\ \\= \frac{3}{2-0}\\ \\= 3/2$

<em>Hence the limit of the function is 3/2.</em>

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3 years ago

PLEASE HELP WILL GIVE BRAINLIEST

defon

Answer:

Thomson ----------- \ /--------------------- Matter is made up of invisible atoms

Rutherford -------------|--|---------------------- Mass of an atom [in the center]

Dalton -------------|-/ /------------------ Electrons move in circular paths...

Bohr -------------|-----|------\ /------ Electrons exist in a cloud of probability.

Schrodinger -------------|----/ \----|------ Atoms contain Electrons

\----------------/

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3 years ago

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