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2 years ago

Which logarithmic equation is equivalent to the exponential equation below e^2x=7

1 answer:

4 0

$\bf log_{{ a}}{{ b}}=y \iff {{ a}}^y={{ b}}\qquad\qquad 
% exponential notation 2nd form
{{ a}}^y={{ b}}\iff log_{{ a}}{{ b}}=y \\\\
-----------------------------\\\\
e^{2x}=7\iff log_e(7)=2x\iff ln(7)=2x$

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What is the missing term?

Nadya [2.5K]

Answer:

3x^2

Step-by-step explanation:

The missing term is the product of x and 3x.

x * 3x = 3x^2

Answer: 3x^2

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3 years ago

Find the final amount of money in an account if $7, 200 is deposited at 2.5 % interest compounded

andreev551 [17]

The final amount is $7,615.27

A = P(1 + r/n)^t

Where,

A = Final amount

P = principal = $7, 200

r = interest rate = 2.5% = 0.025

n = number of periods = 4

t = time = 9 years

A = P(1 + r/n)^t

= 7,200(1 + 0.025/4)^9

= 7,200(1 + 0.00625)^9

= 7,200(1.00625)^9

= 7,200(1.0576769512798)

= 7,615.2740492152

Approximately,

A = $7,615.27

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2 years ago

<img src="https://tex.z-dn.net/?f=%20%5Csqrt%7B%20%7B4%20a%7D%5E%7B2%7D%20%2B%20%20%7B12a%7D%5E%7B2%7D%20%7D%20" id="TexFormula1

PolarNik [594]

Answer:

The answer is 4a

Step-by-step explanation:

1) Simplify 4a² + 12a² to 16a².

$\sqrt{ {16a}^{2} }$

2) Use this rule:

$\sqrt{ab} = \sqrt{a} \sqrt{b}$

$\sqrt{16} \sqrt{ {a}^{2} }$

3) Since 4 × 4 is 16, the square root of 16 is 4.

$4 \sqrt{ {a}^{2} }$

4) Simplify.

$4a$

Therefor, the answer is 4a.

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2 years ago

Find four consecutive integers whose sum is 114.

Nezavi [6.7K]

If n is the first of these integers, then the other three are n + 1, n + 2, and n + 3. Then

n + (n + 1) + (n + 2) + (n + 3) = 114

4n + 6 = 114

4n = 108

n = 27

So the four numbers are 27, 28, 29, and 30.

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2 years ago

1 (5x-3x2-6)-(3x2-5x+7) 2 (2m2+7m+8) -(1-2m-2m2)

swat32

Answer:

#1 -6

#2 14

Step-by-step explanation:

#1 You would put 5x and then you would subtract it by 3 then you multiply it by 3×2 and your answer will be -6

#2 You would add 2 add 7 add 8 and then you would subtract 1 and subtract 2.and your answer 14

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2 years ago