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2 years ago

Using the logarithm, solve the following equation for x. 6400 = (8) (2^4x)

Mathematics

1 answer:

expeople1 [14]2 years ago

8 0

Answer:

x = 2.411

Step-by-step explanation:

6400 = (8) (2^(4x))

Divide both sides by 8 to get;

6400/8 = (2^(4x))

800 = (2^(4x))

Using logarithms, we can solve as;

Log 800 = 4xLog 2

4x = Log800/Log2

4x = 2.9031/0.3010

4x = 9.6449

x = 9.6449/4

x ≈ 2.411

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3 years ago

Wats 3000000000000*30000000000000

Olegator [25]

Answer:

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Step-by-step explanation:

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3 years ago

Read 2 more answers

(-6x3-4x2-8)+(x3-x2-9x+4) add together and simply

Aleksandr [31]

Answer:

-5x³ -5x² -9x - 4

Step-by-step explanation:

(-6x³-4x²-8)+(x³-x²-9x+4)

Add the like terms.

-6x³ and x³ are like terms. -6x³ + x³ = -5x³

-4x² and -x² are like terms. -4x² - x² = -5x²

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3 years ago

When josh borrowed money, he originally agreed to repay the loan by making three equal payments of $1500, with a payment due now

Yuliya22 [10]

Answer:

5 years

Step-by-step explanation:

Principal Amount to be paid=$4500

Interest rate = 2%

Number if Times compounded= number of years

Number of years = x

Among total= $5010

A= p(1+r/n)^(nt)

But n= t =x

A= p(1+r/x)^(x²)

5010=4500(1+0.02/x)^(x²)

5010/4500 = (1+0.02/x)^(x²)

1.11333=( 1+0.02/x)^(x²)

Using trial and error method the number of years maximum to give approximately $5010 is 5 years

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3 years ago

Find a linear second-order differential equation f(x, y, y', y'') = 0 for which y = c1x + c2x3 is a two-parameter family of solu

Alisiya [41]

Let $y=C_1x+C_2x^3=C_1y_1+C_2y_2$ . Then $y_1$ and $y_2$ are two fundamental, linearly independent solution that satisfy

$f(x,y_1,{y_1}',{y_1}'')=0$
$f(x,y_2,{y_2}',{y_2}'')=0$

Note that ${y_1}'=1$ , so that $x{y_1}'-y_1=0$ . Adding $y''$ doesn't change this, since ${y_1}''=0$ .

So if we suppose

$f(x,y,y',y'')=y''+xy'-y=0$

then substituting $y=y_2$ would give

$6x+x(3x^2)-x^3=6x+2x^3\neq0$

To make sure everything cancels out, multiply the second degree term by $-\dfrac{x^2}3$ , so that

$f(x,y,y',y'')=-\dfrac{x^2}3y''+xy'-y$

Then if $y=y_1+y_2$ , we get

$-\dfrac{x^2}3(0+6x)+x(1+3x^2)-(x+x^3)=-2x^3+x+3x^3-x-x^3=0$

as desired. So one possible ODE would be

$-\dfrac{x^2}3y''+xy'-y=0\iff x^2y''-3xy'+3y=0$

(See "Euler-Cauchy equation" for more info)

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3 years ago