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3 years ago

Solve for x if 5^x=200

Mathematics

2 answers:

Sphinxa [80]3 years ago

7 0

Answer:

ln200/ln5

Step-by-step explanation:

ln5^x=ln200

xln5=ln200

x=ln200/ln5

vampirchik [111]3 years ago

5 0

Answer:

x ≈ 3.29

Step-by-step explanation:

Take the logarithm of both sides of the equation to remove the variable from the exponent.

Exact Form:

x = $\frac{ln(200)}{ln(5)}$

Decimal Form:

x = 3.29202967 … ≈ 3.29 (Round to what the question asks, in this case, I rounded to 2 decimal places)

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vovikov84 [41]

$\mathbf f(x,y)=x^4y^5\,\mathbf i+x^5y^4\,\mathbf j$

We want a scalar function $f(x,y)$ whose gradient is equivalent to the vector field. That means

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$\dfrac{\partial f}{\partial y}=x^5y^4\implies x^5y^4=x^5y^4+\dfrac{\mathrm dg}{\mathrm dy}$
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which means (b)

$\displaystyle\int_{\mathcal C}\mathbf f\cdot\mathrm d\mathbf r=f(\mathbf r(1))-f(\mathbf r(0))=f(-1,3)-f(0,0)=-\dfrac{243}5-0=-\dfrac{243}5$

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Step-by-step explanation:

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Solve for x if 5^x=200​

Solve for x if 5^x=200