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I have a Pre-Calculus and Trigonometry final coming up and I need some brushing up on a few concepts that I'm not completely solid on. There are quite a few, but any help is greatly appreciated.

1) Tricky exponential equations where the bases are different and the exponential variable is raised to a power.

I.E. 3^x^3=6^(x+5)

2) Radian word problems where the goal is to convert radians to measurement and then back to radians, given the radii of two circles and the radians turned by one.

I.E. If a pulley of 10-cm radius and a pulley of 14-cm are connected by a rope and the larger pulley turns 6 radians, how many radians with the smaller pulley turn?

3) I need to brush up on finding the answers to trigonometry functions without a calculator since the Final does not allow one.

I.E. sin^-1(2/3)

4) Radical equations with variables in uncombinable roots. I cannot give an example since I can't represent square roots with text.

Thanks in advance. Any help is greatly appreciated.

2007-07-03 12:38:13 · 3 answers · asked by Antonio P 1 in Science & Mathematics Mathematics

Thanks for the reply Amar.

Is there a way to solve a problem like that without a calculator even if it means getting an answer in terms of logarithms? Would an easier equation like 3x^2=6^(x+4)

Thanks again.

2007-07-03 13:39:54 · update #1

Thanks for all of the replies.

The radical question that I was referring to was:

sqrt(5x-3)-2sqrt(x)=6

Thanks for all of the replies. They really help a lot.

After reviewing a bit more, I found two more problems that I was having.

The first problem was with the idea of profit of loss where p is the product price, C is the cost, and R is the revenue.

C = 70 - 11p
R = 14p - 2p^2

I attempted to set the equations equal to each other in order to find the price that equals 0 profit and 0 loss, but the quadratic couldn't be factored.

The other question I had was about frictionless magnitude.

For example, if there was a truck on an incline of 18 degrees and its weight was 6600 pounds, what's the magnitude of the force parallel would be required in order to keep the truck from rolling down the driveway?

2007-07-03 15:02:44 · update #2

3 answers

3^x^3=6^(x+5)
ln 3^x^3 = ln6^(x+5)
x^3 ln 3 = (x+5) ln 6
x^3/(x+5) = ln6/ln3
x^3 -ln6/ln3 x -5ln6/ln3=0
You should be able to solve this from here using your calculator. If calculators are not allowed, you should not be given problems such as this.

2. 2pi radians = 360 degrees
pi radians = 180 degrees
To convert from radians to degrees multiply radians by 180/pi
To convert from degrees to radians multiply degrees by pi/180.

The length of an arc of a circle is s = rx where s is the length of the arc, r is the radius, and x is the radian measure.

Thus if the 14 cm pulley turns 6 radians, the rope passing over the pulley will have a length of 6*14 = 84 cm.

This same amount of rope will pass by the 10 cm pulley. So 10 x =84 --> x = 8.4 radians. Do you see that 14/10= 8.4/6?
The smaller pulley will move Rbig/Rsmall times the radians of the larger pulley.

You should remember the following:
sin 0 =0 sin30=1/2 sin 45= sqrt(2)/2 sin 60=sqrt(3)/2 sin90=1
cos0 =1 cos30=sqrt(3)/2cos45=sqrt(2)/2 cos60=.5 cos90 =0
tan0=0 tan30=sqrt(3)/3 tan45=1 tan60=sqrt(3) tan 90 is undefined. These are the most likely angles to come up. If they ask you to fin arcsin (2/3), you will not be able to do it without a calculator.

Don't understand what you are asking in 4).

2007-07-03 13:41:31 · answer #1 · answered by ironduke8159 7 · 0 1

3^x^3=6^(x+5)
take log of both sides:
x^3 log 3=(x+5)log6
x^3/(x+5=log6/log3)=0.7782/0.4771=1.6311
x^3-1.6311x=5x1.6311=8.155
it is a cubic equation. it is difficult to solve.one root is real&>2 sloghtlyafter finding this real root, the rest involves a quadratic equation which is relatively easy. we need to do numerical work which consumes time and effort. soI leave it here.

2007-07-03 13:10:06 · answer #2 · answered by Anonymous · 0 1

4) Anything in symbols can be represented with text. root(3x + 7) = (3x + 7)^(1/2)

2007-07-03 13:56:12 · answer #3 · answered by Mark 6 · 0 1

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