#1**0 **

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

ElectricPavlov Nov 28, 2018

#2**+1 **

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

Same here. Guest....

By inspection, the only integer solution is x = 0

CPhill Nov 28, 2018

#4**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

ElectricPavlov Nov 28, 2018

#7**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**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

#11**0 **

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

ElectricPavlov
Nov 28, 2018

#13**+2 **

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

HelloWorld
Nov 28, 2018

#14**0 **

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

ElectricPavlov
Nov 28, 2018

#15**+2 **

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

you got from 3^{2(x+1)} + 3^{3(x+1)} = 3^{2} * 4 to 1^{x+1} + 3^{x+1} = 4. Check your answer, you divided the right side by 9 and the left side by (3^{2})^{x+1}=9^{x+1}.

HelloWorld
Nov 28, 2018

#16**0 **

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

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

Thanx for keeping me mathematically honest !

ElectricPavlov
Nov 28, 2018

#17**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

#5**+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

CPhill Nov 28, 2018

#8**+2 **

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

(3^{2})^{x+1} + (3^{3})^{x+1} =(3^{2})^{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

edited by Guest Nov 28, 2018

#19**+10 **

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

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

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

heureka Nov 29, 2018