
Algebra PreAlgebra and Basic Algebra Math Forum 
 LinkBack  Thread Tools  Display Modes 
June 13th, 2017, 02:38 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 04:35 AM. 
June 13th, 2017, 02:56 AM  #2  
Math Team Joined: May 2013 From: The Astral plane Posts: 1,570 Thanks: 613 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
Dan Last edited by skipjack; June 13th, 2017 at 04:50 AM.  
June 13th, 2017, 04:06 AM  #3 
Math Team Joined: Jan 2015 From: Alabama Posts: 2,515 Thanks: 640 
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 04:46 AM. 
June 13th, 2017, 04:20 AM  #4  
Member Joined: Dec 2011 Posts: 61 Thanks: 0  Quote:
Last edited by skipjack; June 13th, 2017 at 04:46 AM.  
June 13th, 2017, 04:33 AM  #5 
Senior Member Joined: Feb 2016 From: Australia Posts: 1,225 Thanks: 420 Math Focus: Yet to find out.  Did you copy the question down correctly? See topsquark's post #2.
Last edited by skipjack; June 13th, 2017 at 04:47 AM. 
June 13th, 2017, 06:05 AM  #6 
Member Joined: Dec 2011 Posts: 61 Thanks: 0  
June 13th, 2017, 06:53 AM  #7 
Senior Member Joined: Apr 2014 From: Glasgow Posts: 2,019 Thanks: 665 Math Focus: Physics, mathematical modelling, numerical and computational solutions  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)  Here's some hints to help you with the mathematics... 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... $\displaystyle 2^7 \div 2^3 = 2^{73} = 2^{4}$ 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! In your equation, ... $\displaystyle 3^x 9^{x+1} = 60$ ... 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 07:09 AM. 
June 22nd, 2017, 10:18 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, 10:35 PM  #9  
Senior Member Joined: Feb 2016 From: Australia Posts: 1,225 Thanks: 420 Math Focus: Yet to find out.  Quote:
 
June 23rd, 2017, 03:49 AM  #10  
Math Team Joined: Jan 2015 From: Alabama Posts: 2,515 Thanks: 640  Quote:
Quote:
Quote:
 

Tags 
equation, indices 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Indices  whiskey tango foxtrot  Algebra  1  January 10th, 2017 07:47 AM 
indices  bongantedd  Algebra  3  January 20th, 2014 10:34 AM 
SOmeone please help me with these indices  Wilsonab  Number Theory  1  November 11th, 2012 02:16 PM 
Indices Help!!  Alexis87  Algebra  11  February 5th, 2012 08:52 AM 
Indices  sallyyy  Algebra  2  August 20th, 2011 05:18 AM 