#### Rengoku0510

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- Thread starter Rengoku0510
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You'll do the same sort of thing you did for (c), only more so. If nothing else, show us your work for (c), so we can use that as a start.Hi, I would like to get some help on question d.

I don't even know where to start

As you presumably did there, you can start by expressing some of the bases as powers of other bases. After all, that's the hint they gave you!

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I like your example, yet I would have finished with[math]= \frac{2^{5x}}{2^{5x}} = 2^{5x-5x} = 2^0 = 1[/math]

[math]= \frac{2^{5x}}{2^{5x}} = 1[/math]

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That's good, Rengoku, with two fixes.

3^9 = 19683 not 27

5^5 = 3125 not 125

Next, use the commutative property to write a ratio of powers of 3 multiplied by a ratio of powers of 5. In other words, swap the numerators. Then, expand the exponents (use distributive property).

See whether that helps you finish.

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ie, \(\displaystyle 2^0=1, 2^1=1, 2^2=4, 2^3=8, 2^4=16, 2^5= 32, 2^6=64, 2^7=128, 2^8=256, 2^9=512, 2^{10}=1024\)

\(\displaystyle 3^0=1, 3^1=3, 3^2=9, 3^3=27, 3^4=81, 3^5=243\)

\(\displaystyle 5^0=1, 5^2=25, 5^3=125, 5^4=625\) for starters.

The first "thing" I see when looking at (d) is that all the bases can be written in terms of 3s and 5s. That's because I

I used to tell my students to write the above on a big piece of paper, paste it up in the toilet, and look at it every day!!

So the first thing to do is replace 27 with (\(\displaystyle 3^3)\) and 125 with (\(\displaystyle 5^3)\) which is what you attempted to do. Brackets are very important.

Now show us what you can do.

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I'll rest my brain for now and try solving this again tomorrow to see if I could do it without getting stuck!

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Good on you for not giving up!! Well done!

I'll rest my brain for now and try solving this again tomorrow to see if I could do it without getting stuck!

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Yes, that's fine.May I post my attempt at this since op has said they have got the answer?

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Sorry for poor quality... Last line says: 5^- 3n-3 * 3^n+2 = 15^-2n-1

When you type this in horizontal style, you must use grouping symbols, as in

5^(-3n - 3)*3^(n + 2).

So the first rule doesn't apply here since they don't have the same base, thanks.

I'm not sure if the second rule applies; does the fact that they both have n as part of their exponents mean they have the same exponents or must it be exactly 3n-3 or n+2?

EDIT:

Question says simplify so do I just leave it as 5^(-3n - 3)*3^(n + 2)? Is it not possible to multiply these then?

Last edited:

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The answer is the LHS of your last line. You can't do anything more. To simplify the expression. It's important to know when to stop.

Question says simplify so do I just leave it as 5^(-3n - 3)*3^(n + 2)? Is it not possible to multiply these then?

The answer is the LHS of your last line. You can't do anything more. To simplify the expression. It's important to know when to stop.

There is an alternative. Write it as the simplified fraction

\(\displaystyle \dfrac{3^{n + 2}}{5^{3n + 3}}. \)

Last edited:

Thanks but how do you know 3^(n+2) is the numerator and not 5^(3n+3)?There is an alternative. Write it as the simplified fraction

\(\displaystyle \dfrac{3^{n + 2}}{5^{3n + 3}}. \)

Thanks but how do you know 3^(n+2) is the numerator and not 5^(3n+3)?

If you have a positive leading coefficient for the exponent, then the corresponding term goes in

the numerator and vice versa.

See that 5^(-3n - 3) from the product is equal to

5^[-(3n + 3)]. This is equivalent to 1/[5^(3n + 3)].

I must go now.