Solve the equation

Question

Solve the equation (2 · 3^x)³ + (9x − 3)³ = (9^x + 2 · 3^x − 3)³

Answer 1

I will answer assuming your 9x term is meant to be 9^x. Please reply if that is not the case.

If you make the substitutions 2·3^x=a and 9^x-3=b, we can write your equation more cleanly as

a³+b³=(a+b)³

Now we can expand the right-hand side and see the cancellation more easily. The final expansion is

a³+b³=a³+3a²b+3ab²+b³

Cancelling like terms yields:

0=3a²b+3ab²

Divide by 3 and factor:

0=ab(a+b)

So either a=0, b=0, or a+b=0.

a=0 would mean 3^x=0, which never happens. We get no solutions there.

b=0 would mean 9^x-3=0

9^x=3

(3²)^x=3

3^2x=3¹

2x=1

x=1/2

And a+b=0 would mean

2·3^x+9^x-3=0

2·3^x+3^2x-3=0

(3^x)²+2·3^x-3=0

This is a quadratic in 3^x, so we apply the quadratic formula

3^x=-2±√(2²-4·1·(-3)) /2·1

3^x=-2±√16 /2

3^x=-2±4 /2

3^x=-1±2

3^x=1 or -3

3^x is never negative, so 3^x=1, or x=0.

So the solutions are x=0 and x=1/2.

Answer 2

This is a transcendental equation (meaning it has both "x" as a base and "x" as an exponent) and is not easily solved using algebraic methods. The easiest way to proceed is to use graphical methods like your graphing calculator or desmos.

Enter one equation as y = (2•3^x)³ + (9x - 3)³ and enter the second equation as y = (9^x + 2•3^x - 3)³ then look for the intersection points of the 2 graphs.

If you need help with that, let me know.