+36916 I am just looking t it and can see x = 0....I'll let Chris work out the details.....
+128448 3^(2x+2)+27^(x+1)=36
Same here. Guest....
By inspection, the only integer solution is x = 0
+36916 = 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
+36916 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!
+36916 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
+36916 Sure you can.....let's call it "divide both sides of the equation by 9 '
+25 EP on the right side you divide by 9 and on the left side you divide by 9x+1, you can't do that
+36916 No.....I divided by 9 then I simplified 1^(x+1) to ' 1 ' since 1 to any power is still just 1 
+25 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.
+36916 NOW I see what you are saying......
working on yet ANOTHER (in?)correct solution...
Thanx for keeping me mathematically honest !
+36916 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....
+128448 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
+25 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)
+26367 exponent problem
solve
\(x\) for \(\large 3^{(2x+2)}+27^{(x+1)}=36\)
\(\text{Formula:}\\ \large{\boxed{\left(a^b\right)^c = a^{bc}=a^{cb}=\left(a^c\right)^b}}\)
\(\begin{array}{|rcll|} \hline 3^{(2x+2)}+27^{(x+1)} &=& 36 \quad | \quad 27 = 3^3 \\ 3^{2( x+1)}+\left(3^3\right)^{(x+1)} &=& 36 \quad | \quad \left(3^3\right)^{(x+1)} = 3^{3(x+1)}\\ 3^{2( x+1)}+ 3^{3(x+1)} &=& 36 \\ \left(3^{(x+1)}\right)^{2} + \left(3^{(x+1)}\right)^{3} &=& 36 \\ && \text{we substitute:} ~ \boxed{u = 3^{(x+1)}} \\\\ u^2+u^3&=& 36 \\ u^3+u^2-36 &=& 0 \\\\ u_1 &=& 3 \\ u_2 &=& -2-2i\sqrt{2} \\ u_3 &=& -2+2i\sqrt{2} \\\\ u &=& 3^{(x+1)} \quad | \quad u = 3 \\ 3 &=& 3^{(x+1)} \\ 3^1 &=& 3^{(x+1)} \\ 1 &=& x+1 \\ \mathbf{ x } & \mathbf{=} & \mathbf{0} \\ \hline \end{array}\)
