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I think 5 is raising to the power nothing therefore should be equal to zero and not one.
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Supposing you had
2^4 * 2^0
when we multiply the same bases we are allowed to add the exponents. 4 + 0 = 4 so 2^4 * 2^0 = 2^4 so 2^0 must equal 1.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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But is it possible to add the exponents of the following?
Last edited by EbenezerSon (2013-07-23 23:45:13)
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No, it is not. They do not have the same base.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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It is due to such situations why I could not solve the previous problem since it has 6*3^2n+3. the six made it hard for me to manipulate. One cannot reduce it to be three. Or if the six had been three I would multiply it with the three to get nine and then reduce it to three, so I could manipulate. Since nine would be a perfect square.
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Then the final answer would be 9*16 = 144, is that right?
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6 = 2 * 3 so you could have combined that statement into
6*3^(2n+3) = 2 *3 * 3^(2n+3)= 2 * 3^(2n+4)
and you are done.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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6 = 2 * 3 so you could have combined that statement into
6*3^(2n+3) = 2 *3 * 3^(2n+3)= 2 * 3^(2n+4)
and you are done.
I don't understand yours, I thought it would be
2*3*3(2n+3) = 2*9^(2n+3) = 2*3^3(2n+3) = 2*3^(6n+9)
Last edited by EbenezerSon (2013-07-24 00:20:56)
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I must review your procedure with the previous problem to comprehend.
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You add coefficients:
2 * 3^1 * 3^(2n+3)
2n +3 + 1 = 2n + 4 so
2 * 3^(2n+4)
Are you saying that
because that is incorrect.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Errrr okay I have got you, then what would one make of the two, I mean the base.
Last edited by EbenezerSon (2013-07-24 00:33:59)
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You add coefficients:
2 * 3^1 * 3^(2n+3)
2n +3 + 1 = 2n + 4 so
2 * 3^(2n+4)
Are you saying that
because that is incorrect.
The two I have underlined above.
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Hi;
You can only do this:
6 * 3^(2n+3)
2 * 3 * 3^(2n+3)
2 * 3^1 * 3^(2n+3)
Add the exponents on the threes.
2n +3 + 1 = 2n + 4 so
2 * 3^(2n+4)
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Okay I digest it. But can I use that procedure to arrive on that answer 7 you had in the previous problem?
Last edited by EbenezerSon (2013-07-24 00:44:58)
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That problem is a real bear and I recommend the other method, the one I posted.
I am going to get a little bit of sleep see you later.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Okay, Thanks God bless. I shall solve more problems.
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How did you do?
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Hi, bobbym I have come across another confusing one with different bases.
Solve for X.
3^x * 2^x-1 = 1.
How would one go about this? Is wierd.
Last edited by EbenezerSon (2013-07-24 04:01:51)
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Is that
because if it is...
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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The book has 0.39 answer
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That is correct if you round my answer to 2 decimal places.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Is that
because if it is...
Please help me understand this.
Thanks in advance
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I can show you the steps but unless you have a little knowledge about logarithms it will be confusing.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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I had solved much problems on logarithms.
Usually I had encountered problems on indices with unequal bases, and the book used log to solve but I can't percieve this would need logarithm application.
Please carry on!
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Okay, I will provide the steps:
Take the log of both sides.
Add log(2) to both sides.
Divide both sides by ( log(2) + log(3) ).
And we are done.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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