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 June 13th, 2017, 03:38 AM #1 Member   Joined: Dec 2011 Posts: 61 Thanks: 0 indices equation hai. Can you show me how to solve this equation? Really appreciate it. (3x)(9^{x+1})= 60 Last edited by skipjack; June 13th, 2017 at 05:35 AM.
June 13th, 2017, 03:56 AM   #2
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Quote:
 Originally Posted by najaa hai. Can you show me how to solve this equation? Really appreciate it. (3x)(9^{x+1})= 60
As written, it can't be done. Did you perhaps mean $\displaystyle (3^x)(9^{x + 1}) = 60$?

-Dan

Last edited by skipjack; June 13th, 2017 at 05:50 AM.

 June 13th, 2017, 05:06 AM #3 Math Team   Joined: Jan 2015 From: Alabama Posts: 2,922 Thanks: 785 It can be done, just not with "elementary" functions: $\displaystyle 3x(9^{x+1})= 27(x9^x)= 60$ $\displaystyle x9^x= \frac{60}{27}= \frac{20}{9}$ $\displaystyle xe^{x\ln(9)}= \frac{20}{9}$. Let $y= x\ln(9)$ so $x= y/\ln(9)$. The equation becomes $\displaystyle \frac{ye^y}{\ln(9)}= \frac{20}{9}$ so that $\displaystyle ye^y= \frac{20\ln(9)}{9}$. Now apply "Lambert's W function" (defined as the inverse function to $\displaystyle f(x)= xe^x$) to both sides: $\displaystyle y= W\left(\frac{20 \ln(9)}{9}\right)$ so that $\displaystyle x= \frac{y}{\ln(9)}= \frac{W\left(20 \ln(9)\right)}{9 \ln(9)}$. Yes, $\displaystyle 3^x 9^{x+ 1}= 60$ is much simpler. $\displaystyle 9= 3^2$ so we can write the equation as $\displaystyle 9(3^x)(3^{2x})= 9*3^{3x}= 60$ $\displaystyle 3^{3x}= \frac{60}{9}= \frac{20}{3}$. Last edited by skipjack; June 13th, 2017 at 05:46 AM.
June 13th, 2017, 05:20 AM   #4
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Quote:
 Originally Posted by Country Boy It can be done, just not with "elementary" functions: $\displaystyle 3x(9^{x+1})= 27(x9^x)= 60$ $\displaystyle x9^x= \frac{60}{27}= \frac{20}{9}$ $\displaystyle xe^{x\ln(9)}= \frac{20}{9}$. Let $y= x\ln(9)$ so $x= y/\ln(9)$. The equation becomes $\displaystyle \frac{ye^y}{\ln(9)}= \frac{20}{9}$ so that $\displaystyle ye^y= \frac{20\ln(9)}{9}$. Now apply "Lambert's W function" (defined as the inverse function to $\displaystyle f(x)= xe^x$) to both sides: $\displaystyle y= W\left(\frac{20 \ln(9)}{9}\right)$ so that $\displaystyle x= \frac{y}{\ln(9)}= \frac{W\left(20 \ln(9)\right)}{9 \ln(9)}$. Yes, $\displaystyle 3^x 9^{x+ 1}= 60$ is much simpler. $\displaystyle 9= 3^2$ so we can write the equation as $\displaystyle 9(3^x)(3^{2x})= 9*3^{3x}= 60$ $\displaystyle 3^{3x}= \frac{60}{9}= \frac{20}{3}$.
Sorry, so what's the final answer? I don't get it.

Last edited by skipjack; June 13th, 2017 at 05:46 AM.

June 13th, 2017, 05:33 AM   #5
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Quote:
 Originally Posted by najaa Sorry, so what's the final answer? I don't get it.
Did you copy the question down correctly? See topsquark's post #2.

Last edited by skipjack; June 13th, 2017 at 05:47 AM.

June 13th, 2017, 07:05 AM   #6
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Quote:
 Originally Posted by Joppy Did you copy the question down correctly? See topsquark's post #2.
Yes I think so

June 13th, 2017, 07:53 AM   #7
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Quote:
 Originally Posted by najaa Yes I think so
If you're struggling with this, chances are you don't understand index laws.

I'm assuming your problem is this:

Solve $\displaystyle 3^x 9^{x+1} = 60$

NOT this...

Solve $\displaystyle 3x \times 9^{x+1} = 60$

because that latter equation is much much more difficult to solve (as per Country Boy's solution with a Lambert W function)

-----

First... A little bit about notation.

In your problem above, you have some quantities which are superscript. Superscript is the name for any text which is written in a slightly smaller font floating a little bit above another thing. For example, the 3 in $\displaystyle 2^3$ is superscript. In mathematics, if something has some superscript to the right of the number, it denotes an "exponent". The exponent represents how many times a number is multiplied by itself and is sometimes called a "power" or an "index". The number that has the exponent is called the "base". Therefore, if you have $\displaystyle 2^3$, the exponent is 3 and the base is 2 ($\displaystyle base^{exponent}$).

Second: Index laws

Index laws are equations that relate quantities together that have different exponents. All index laws require the same base to work.

Law 1: if you multiply, add the powers.

For example... $\displaystyle 2^3 \times 2^7 = 2^{3+7} = 2^{10}$

Law 2: if you divide, subtract the powers

For example... $\displaystyle 2^7 \div 2^3 = 2^{7-3} = 2^{4}$

Law 3: if you have a power of a power, multiply

For example... $\displaystyle (2^7)^3 = 2^{7\times 3} = 2^{21}$

Law 4: anything to a power 0 is 1.

For example... $\displaystyle (7895789473895643^{543})^0 = 1$

It doesn't matter what is in the bracket *
* Caveat... if you get $\displaystyle 0^0$... fun stuff happens. We won't worry about that! *cough*

Third... solving equations.

Top tip: Remember that the first two index laws require the same base. Super important!

$\displaystyle 3^x 9^{x+1} = 60$

... you have a number with a base of 3 and an exponent of x multiplied by a number with a base of 9 and an exponent of x+1. We can't do anything with these yet because the bases are different.

Therefore, let's make the bases the same!

It turns out that 9 is equal to $\displaystyle 3^2$. That's fortunate!

Note: You'll probably be given lots of questions that involve squares (1, 4, 9, 16, 25, 36, 49, etc.) and cubes (1, 8, 27, 64, 216, etc.) so you can do this kind of substitution. Revise those lists because they'll make the problems much easier to solve

We can try and describe the second bit of the equation using $\displaystyle 3^2$. Let's substitute it in:

$\displaystyle 3^x 9^{x+1} = 60$
$\displaystyle 3^x (3^2)^{x+1} = 60$

Using index law 3, we get

$\displaystyle 3^x 3^{2(x+1)} = 60$

We now have two quantities with the same base... that means we can use the other index laws now.

Since we're multiplying the $\displaystyle 3^x$ and the $\displaystyle 3^{2(x+1)}$, we use index law 1 to get

$\displaystyle 3^{x + 2(x+1)} = 60$

Can you finish it from here?

Hint... You first simplify the exponent using algebra, then rearrange for x using laws of logs.

Give us a shout if you get stuck.

Last edited by Benit13; June 13th, 2017 at 08:09 AM.

 June 22nd, 2017, 11:18 PM #8 Newbie   Joined: Apr 2017 From: durban Posts: 10 Thanks: 0 Math Focus: Algebra ... you have a number with a base of 3 and an exponent of x multiplied by a number with a base of 9 and an exponent of x+1. We can't do anything with these yet because the bases are different. Therefore, let's make the bases the same! It turns out that 9 is equal to 32. That's fortunate! Note: You'll probably be given lots of questions that involve squares (1, 4, 9, 16, 25, 36, 49, etc.) and cubes (1, 8, 27, 64, 216, etc.) so you can do this kind of substitution. Revise those lists because they'll make the problems much easier to solve We can try and describe the second bit of the equation using 32. Let's substitute it in: 3x9x+1=60 3x(32)x+1=60 Using index law 3, we get 3x32(x+1)=60 We now have two quantities with the same base... that means we can use the other index laws now. Since we're multiplying the 3x and the 32(x+1), we use index law 1 to get 3x+2(x+1)=60 Nice analysis thet just d right way solve an Exponential problem.
June 22nd, 2017, 11:35 PM   #9
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Quote:
Originally Posted by Ola Lawson
Quote:
 Originally Posted by Benit13 ... you have a number with a base of 3 and an exponent of x multiplied by a number with a base of 9 and an exponent of x+1. We can't do anything with these yet because the bases are different. Therefore, let's make the bases the same! It turns out that 9 is equal to 32. That's fortunate! Note: You'll probably be given lots of questions that involve squares (1, 4, 9, 16, 25, 36, 49, etc.) and cubes (1, 8, 27, 64, 216, etc.) so you can do this kind of substitution. Revise those lists because they'll make the problems much easier to solve We can try and describe the second bit of the equation using 32. Let's substitute it in: 3x9x+1=60 3x(32)x+1=60 Using index law 3, we get 3x32(x+1)=60 We now have two quantities with the same base... that means we can use the other index laws now. Since we're multiplying the 3x and the 32(x+1), we use index law 1 to get 3x+2(x+1)=60
Nice analysis thet just d right way solve an Exponential problem.
You can use the [QUOTE] tags to distinguish other peoples comments from your own.

June 23rd, 2017, 04:49 AM   #10
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Quote:
 Originally Posted by Ola Lawson ... you have a number with a base of 3 and an exponent of x multiplied by a number with a base of 9 and an exponent of x+1. We can't do anything with these yet because the bases are different. Therefore, let's make the bases the same! It turns out that 9 is equal to 32. That's fortunate!
No, 9 is not equal to 32. It's equal to $9^2$ or, if you don't want to use Latex, 3^2.

Quote:
 Note: You'll probably be given lots of questions that involve squares (1, 4, 9, 16, 25, 36, 49, etc.) and cubes (1, 8, 27, 64, 216, etc.) so you can do this kind of substitution. Revise those lists because they'll make the problems much easier to solve We can try and describe the second bit of the equation using 32. Let's substitute it in: 3x9x+1=60 3x(32)x+1=60
The way this is written is non-sense. You mean 3^x(9^{x+ 1})= 60 and (3^x)(3^2)^(x+ 1)= 60.

Quote:
 Using index law 3, we get 3x32(x+1)=60 We now have two quantities with the same base... that means we can use the other index laws now. Since we're multiplying the 3x and the 32(x+1), we use index law 1 to get 3x+2(x+1)=60 Nice analysis thet just d right way solve an Exponential problem.
What in the world are "index law 1" and "index law 3"? Do you think that everyone in the world learns the "index laws" in the same order? (or even calls them "index laws"? To me an "index" is just a subscript or superscript used to distinguish between similar quantities. Here we are talking about "exponents" and using "exponential laws".)

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