+0  
 
0
385
20
avatar

solve x for 3^(2x+2)+27^(x+1)=36

 Nov 28, 2018
 #1
avatar+20239 
0

I am just looking t it and can see x = 0....I'll let Chris work out the details.....

 Nov 28, 2018
 #2
avatar+107405 
+1

 3^(2x+2)+27^(x+1)=36

 

Same here. Guest....

 

By inspection, the only integer solution is   x = 0

 

 

cool cool cool

 Nov 28, 2018
edited by CPhill  Nov 28, 2018
 #3
avatar
+1

i want to know solving processes

Guest Nov 28, 2018
 #4
avatar+20239 
0

= 3^(2(x+1)) + 3^(3(x+1)) = 36

(3^2 +3^3)^(x+1) = 36                           this is incorrect!!!  See CPhil answer below!

36^(x+1) = 36

(x+1) Log 36 = log 36

x+1 = 1

x=0

 Nov 28, 2018
edited by ElectricPavlov  Nov 28, 2018
 #6
avatar+107405 
0

Heck, EP.....you beat me to it....LOL!!!!

 

cool cool cool

CPhill  Nov 28, 2018
 #7
avatar+20239 
0

Hmmmm......but I think I made an error that you didn't !    Step 1 to step 2 is not true (now in red).....but the answer worked out by chance!

ElectricPavlov  Nov 28, 2018
 #9
avatar+20239 
0

Here is a better solution:

 

3^2(x+1) + 3^(3(x+1) = 3^2 * 4    Divide by 3^2

1^(x+1) + 3^(x+1) = 4

1 + 3^(x+1) = 4

3^(x+1) = 3

(x+1) LOG 3 = LOG 3

x+1 = 1

x=0

ElectricPavlov  Nov 28, 2018
 #10
avatar+25 
+2

You can't divide by 32 like that

HelloWorld  Nov 28, 2018
 #11
avatar+20239 
0

Sure you can.....let's call it "divide both sides of the equation by 9 '

ElectricPavlov  Nov 28, 2018
 #13
avatar+25 
+2

EP on the right side you divide by 9 and on the left side you divide by 9x+1, you can't do that

HelloWorld  Nov 28, 2018
 #14
avatar+20239 
0

No.....I divided by 9     then I simplified  1^(x+1)   to ' 1 '   since 1 to any power is still just 1  cheeky

ElectricPavlov  Nov 28, 2018
 #15
avatar+25 
+2

the equation is 32x+2+27x+1=36, not 3*(2x+2)+27*(x+1)=36. 

 

you got from 32(x+1) + 33(x+1) = 32 * 4 to 1x+1 + 3x+1 = 4. Check your answer, you divided the right side by 9 and the left side by (32)x+1=9x+1.

HelloWorld  Nov 28, 2018
edited by HelloWorld  Nov 28, 2018
edited by HelloWorld  Nov 28, 2018
edited by HelloWorld  May 15, 2019
 #16
avatar+20239 
0

NOW I see what you are saying......

   working on yet ANOTHER (in?)correct solution...

       Thanx  for keeping me mathematically honest !  cheeky

ElectricPavlov  Nov 28, 2018
 #17
avatar+20239 
0

I can see a graphical solution shows x= 0...and I can see it by LOOKING at it....but I am kinda stuck as to how to solve it algebraically...   Hmmmmm....

ElectricPavlov  Nov 28, 2018
 #18
avatar+25 
+2

No problem laugh

HelloWorld  Nov 28, 2018
 #5
avatar+107405 
+1

3^(2x + 2)  + 27^(x + 1)  = 36      we can write

 

[ 3^2] ^(x + 1)  + [3^3]^(x + 1)] = 36       factor the left side

 

[3^2]^(x + 1) [ 1 + 3] = 36

 

[3^2]^(x + 1) [4] = 36       divide both sides by 4

 

[3^2]^(x + 1) = 9   

 

[9]^(x + 1) = 9^1

 

Since the bases are the same, solve for the exponents

 

x + 1    = 1      subtract 1 from both sides

 

x = 0

 

 

cool cool cool

 Nov 28, 2018
edited by CPhill  Nov 28, 2018
 #8
avatar+25 
+2

I don't think your factoring is correct, the equation

 

(32)x+1 + (33)x+1 =(32)x+1*(1+3)

 

Is not always true (although the only solution is x=0)

HelloWorld  Nov 28, 2018
edited by Guest  Nov 28, 2018
edited by Guest  Nov 28, 2018
 #12
avatar+107405 
0

Yeah...I see my error....!!!

 

You are correct....!!!

 

 

cool cool cool

CPhill  Nov 28, 2018
edited by CPhill  Nov 28, 2018
 #19
avatar+24093 
+10

exponent problem
solve
\(x\)  for  \(\large 3^{(2x+2)}+27^{(x+1)}=36\)

 

\(\text{Formula:}\\ \large{\boxed{\left(a^b\right)^c = a^{bc}=a^{cb}=\left(a^c\right)^b}}\)

 

\(\begin{array}{|rcll|} \hline 3^{(2x+2)}+27^{(x+1)} &=& 36 \quad | \quad 27 = 3^3 \\ 3^{2( x+1)}+\left(3^3\right)^{(x+1)} &=& 36 \quad | \quad \left(3^3\right)^{(x+1)} = 3^{3(x+1)}\\ 3^{2( x+1)}+ 3^{3(x+1)} &=& 36 \\ \left(3^{(x+1)}\right)^{2} + \left(3^{(x+1)}\right)^{3} &=& 36 \\ && \text{we substitute:} ~ \boxed{u = 3^{(x+1)}} \\\\ u^2+u^3&=& 36 \\ u^3+u^2-36 &=& 0 \\\\ u_1 &=& 3 \\ u_2 &=& -2-2i\sqrt{2} \\ u_3 &=& -2+2i\sqrt{2} \\\\ u &=& 3^{(x+1)} \quad | \quad u = 3 \\ 3 &=& 3^{(x+1)} \\ 3^1 &=& 3^{(x+1)} \\ 1 &=& x+1 \\ \mathbf{ x } & \mathbf{=} & \mathbf{0} \\ \hline \end{array}\)

 

laugh

 Nov 29, 2018
edited by heureka  Nov 29, 2018
 #20
avatar+107405 
+1

Thanks, heureka......!!!

 

[ I think we tried to make it more complicated than need be...!!!]

 

 

cool cool cool

CPhill  Nov 29, 2018
edited by CPhill  Nov 30, 2018

25 Online Users

avatar