To solve exponential equations like 3^x + 9^x + 27^x = 39, express all terms with the same base (3^x + 3^(2x) + 3^(3x) = 39), substitute y = 3^x to transform it into a polynomial equation (y + y² + y³ = 39), rearrange to form y³ + y² + y - 39 = 0, factor using algebraic identities (difference of squares and cubes) to get (y-3)(y² + 4y + 13) = 0, apply the zero product rule to find y = 3, then solve for x by recognizing that 3^x = 3 implies x = 1.
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Olympiad Mathematics | Japanese | Can You Solve This New One?Added:
If you're ready, let's provide the solution to this one very quickly.
We have 3 to the^ x + 9 ^ x + 27 ^ x = 39.
Okay. So, what we're going to do here is to simplify what we have here.
Okay, remember 9 is 3 squar and that is to the power of x. Then + 27 is 3 cube and that is still to the^ of x. So this is equal to 39.
And now remember that the position of the powers here can be changed because if you have a to the power of b to the power of c, this can be written as a to the power of c then to the power of b. So the one that comes first does not really matter.
So the same thing will happen to this powers as we have 3 ^ x + 3 ^ x and the square goes outside plus here we have 3 ^ x and the cube goes outside.
This is all equal to 39.
Now we have 3 ^ x in three places. So let letter let y be = 3 to the^ of x and that means that we'll be having here y plus here we have y 2 and then here we have y have y cub and it's equal to 39 and um we cannot add the left hand side but let's look at 39. We can break 39.
Yes, if we break it, we are going to have y + y 2 + y cub to be equal to 39 is going to be 3 + 9 + 27.
Yes, 3 + 9 + 27. And then this is y + y 2 + y cub to be equal to 3 + 3 2 + y 2 y cub um 3 cube rather that is for the 27 3 cube and from here we are going to reposition what we have Y and 3 will come together y - 3.
Okay. Then we have + this is y^ 2 and 3 y. So it's becomes sorry 3 2 becomes - 3^ 2. This will be together.
Then + y cube and 3 cube together.
This is cube and everything is now equal to zero since we have moved everything to the left. Now let us express um this and this as difference of two squares and difference of two cubes respectively.
So this will come down we have y - 3.
Okay. Then plus here we have difference of two squares and that's going to be y - 3 * y + 3.
This is for the difference of two squares.
And then we go over to the difference of two cubes from here. And that will be open bracket. We have y - 3 * y^ 2 + 3 y then + 3 2 and 3 2 is 9.
So this is all equal to zero. So what I expressed over there is what we call the difference of two cubes.
Remember if you have a cub - b cube.
This is equal to a uh minus b * a 2 + a b + b 2. Okay. And this is the identity.
This is the identity that gave me the difference of two cubes here. Okay. So I will remove this and then we'll continue.
Now, if you look at what we have here, y - 3 is common to the three of them. So, I'm going to pick out y - 3 as the common factor. Then we open bracket. y - 3 / itself is 1 plus here we have y + 3.
y + 3 then plus this is already out. So here we have y^ 2 + 3 y + 9 and then we equate to zero.
Okay, we equate to zero like that. And then we have y - 3 still the common factor.
And uh we have I want to write this one first. y² should come down first. Then y + 3 y that will give 4 y.
Then I have um 1 + 3 that is 4. 4 + 3 uh 9 that will give 13.
So we have 13 and this is equal to zero. So at this point we apply our zero product rule saying that it is either this is equal to zero or this is equal to z. But if we pick 1 - 3 to be zero, then it means that it means that y is equal to 3. Now I will come back to this. Y is already equal to 3. Now let's go back to this expression and equate to zero.
Okay. So, we have y 2 we have y ^2 + 4 y + 13 = um 0.
And we're going to use um the quadratic formulas to work this. Right? The formula is y = - b + or minus square<unk> b 2 - 4 a c all over 2 * a. Now let's see if this will give a real solution. So y is going to be minus b which is now minus 4 because b is 4 plus or minus we have the square root of b ^ 2 which is 4^ 2 - 4 * 1 cuz the square root of um the coefficient of y^ 2 is is 1 and that is a then multiply by 13.
Okay. So this is all over 2 * 1. So from here now our y is going to be -4 + or minus we have 4 2 is 16 right and 4 * 13 that will give 52.
So we write 52 over there and this is all over 2 * 1 which is 2.
Okay. So from here let's continue.
Okay. And if we go on, y is going to be -4 + or minus roo<unk> of 36 / 2. Now we are going to have square t and because of that whatever value of y we are going to have from here will be rejected.
Okay. So we are going to reject whatever value of y from here. So this means that we are having only one value of y.
Yes. And it is our y to be equal to 3.
But initially there was no y in the equation.
And you can remember when we say that y should be equal to 3 ^ x. So 3 ^ x is now equal to 3 meaning that 3 here can have the power one. So this means that the value of x is equal to 1 because if the bases are equal then the powers are equal. So the only value of x that satisfies the equation is x = 1.
Thank you for watching.
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