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3 years ago

NEED HELP NOW!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Mathematics

1 answer:

Contact [7]3 years ago

3 0

<span>which means the word ur</span>

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How to solve equations in which the variable is in an exponent

Temka [501]

Answer:

It depends on the equation.

If the bases are equal and the variables are only in the exponents, set the exponents equal.

If there are variables in the exponents, but you cannot set the bases equal, then use logarithms.

Example 1:

$4^{2x + 5} = 4^9$

Here you have the same base on both sides. The variables are in the exponents. Set the exponents equal and solve for x.

2x + 5 = 9

2x = 4

x = 2

Example 2:

$3^{2x} = 9^{6}$

The bases are different, but you can make the bases equal using laws of exponents. Remember that 9 = 3^2.

$3^{2x} = (3^2)^{6}$

$3^{2x} = 3^{12}$

Now you have equal bases, so the exponents must be equal.

2x = 12

x = 6

Example 3:

$10^{x + 2} = 9$

Here you can't make the bases equal, so you take the log of both sides and use laws of logs.

$\log(10^{x + 2}) = \log 3^2$

$(x + 2) \log 10 = 2 \log 3$

Recall that: $\log 10 = 1$

$x + 2 = 2 \log 3$

$x = 2 \log 3 - 2$

3 0

3 years ago

I have another question as well!

vampirchik [111]

You need to simplify 3(2x-5) which would make it 6x-15 on the bottom.
Then take 4x + 3 in and multiply it by 6x - 15 to get the area.

PERIMETER= 8x + 6 + 12x - 30
= 20x - 24

5 0

3 years ago

Use Stokes' Theorem to evaluate S curl F · dS. F(x, y, z) = 5y cos(z) i + ex sin(z) j + xey k, S is the hemisphere x2 + y2 + z2

Alenkinab [10]

By Stokes' theorem, the integral of the curl of $\vec F$ across $S$ is equal to the integral of $\vec F$ along the boundary of $S$ , call it $C$ . Parameterize $C$ by

$\vec r(t)=2\cos t\,\vec\imath+2\sin t\,\vec\jmath$

with $0\le t\le2\pi$ . So we have

$\displaystyle\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S=\int_C\vec F\cdot\mathrm d\vec r$

$=\displaystyle\int_0^{2\pi}(10\sin t\cos 0\,\vec\imath+e^{2\cos t}\sin0\,\vec\jmath+2\cos t\,e^{2\sin t}\,\vec k)\cdot(-2\sin t\,\vec\imath+2\cos t\,\vec\jmath)\,\mathrm dt$