Math Help - exponential equation

1. exponential equation

9^(x+1) + 3^(x+2) - 4 = 0

can anyone show me how to solve this.

2. Originally Posted by fabregas99
9^(x+1) + 3^(x+2) - 4 = 0

can anyone show me how to solve this.
Since $9^{x+1} = 9^x \cdot 9$ and $9 = 3^2$ it follows that $9^{x+1} = 9 \cdot (3^2)^x = 9 \cdot (3^x)^2$.

Furthermore, $3^{x+2} = 3^x \cdot 3^2 = 9 \cdot 3^x$.

Therefore your equation can be re-written as

$9 \cdot (3^x)^2 + 9 \cdot 3^x - 4 = 0 \Rightarrow (3 \cdot 3^x + 4) (3 \cdot 3^x - 1) = 0$.

3. thanks

i've tried working it out, but i cant get the answer that it has in the back of the practice exam. which is

x=log3(4/3)

(thats log to the base 3 couldnt find subscript)
9^(x+1) = 3^(x+3)
so
3^(x+3) - 3^(x+2) -4 = 0

then you can take logs to the base 3 of the first 2 exponentials but when it come to 4, i relaly dont understand what to do. reall sorry on my first post just realised i put a + when it should have been a - imbetween 9^(x+1) and 3^(x+2). can u explain.

4. Originally Posted by fabregas99
i've tried working it out, but i cant get the answer that it has in the back of the practice exam. which is

x=log3(4/3)

(thats log to the base 3 couldnt find subscript)
9^(x+1) = 3^(x+3)
so
3^(x+3) - 3^(x+2) -4 = 0

then you can take logs to the base 3 of the first 2 exponentials but when it come to 4, i relaly dont understand what to do. reall sorry on my first post just realised i put a + when it should have been a - imbetween 9^(x+1) and 3^(x+2). can u explain.
Then you make a small change to what I posted - the equation can be re-written as

$9 \cdot (3^x)^2 - 9 \cdot 3^x - 4 = 0 \Rightarrow (3 \cdot 3^x - 4) (3 \cdot 3^x + 1) = 0$.

So one of the two bracketed expressions is equal to zero ....

5. Originally Posted by fabregas99
9^(x+1) + 3^(x+2) - 4 = 0
Use the rules for exponents to restate in the manner outlined in a previous reply:

. . . . . $9^{x+1}\, =\, 9^x\, \times\, 9^1\, =\, (9^x)(9)\, =\, 9\, \times\, 9^x$

. . . . . $3^{x+2}\, =\, 3^x\, \times\, 3^2\, =\, 3^x\, \times\, 9\, \, 9\, \times\, 3^x$

This makes the equation the same as:

. . . . . $9(9^x)\, +\, 9(3^x)\, -\, 4\, =\, 0$

Then note that 9 = 3^2, so 9^x = (3^2)^x = (3^x)^2, which means the equation is really a "quadratic" in 3^x:

. . . . . $9(3^x)^2\, +\, 9(3^x)\, -\, 4\, =\, 0$

Using the factoring provided earlier, you should arrive at:

. . . . . $3^x\, =\, -\frac{4}{3}\, \mbox{ and }\, 3^x\, =\, \frac{1}{3}$

From what you've learned about exponential functions, you know that you cannot, by powers, turn a positive three into a negative value, so the first of the two equations above has "no solution".

Apply logs or the definition to solve the other exponential equation.

Note: The solution provided cannot be correct. If x = log_3(4/3), then log definitions and log rules give::

. . . . . $x\, +\, 1\, =\, \log_3(4/3)\, +\, \log_3(3)\, =\, \log_3(4)$

. . . . . $x\, +\, 2\, =\, \log_3(4/3)\, +\, \log_3(9)\, =\, \log_3(8)$

. . . . . $9^{x+1}\, =\, \left(3^{\log_3(4)}\right)^2\, =\, (4)^2\, =\, 16$

. . . . . $3^{x+2}\, =\, 3^{\log_3(8)}\, =\, 8$

. . . . . $9^{x+1}\, +\, 3^{x+2}\, -\, 4\, =\, 16\, +\, 8\, -\, 4\, =\, 20$

But the left-hand side is supposed to equal zero.

You might want to point out this typo to your instructor.

6. Originally Posted by stapel
Use the rules for exponents to restate in the manner outlined in a previous reply:

. . . . . $9^{x+1}\, =\, 9^x\, \times\, 9^1\, =\, (9^x)(9)\, =\, 9\, \times\, 9^x$

. . . . . $3^{x+2}\, =\, 3^x\, \times\, 3^2\, =\, 3^x\, \times\, 9\, \, 9\, \times\, 3^x$

This makes the equation the same as:

. . . . . $9(9^x)\, +\, 9(3^x)\, -\, 4\, =\, 0$

Then note that 9 = 3^2, so 9^x = (3^2)^x = (3^x)^2, which means the equation is really a "quadratic" in 3^x:

. . . . . $9(3^x)^2\, +\, 9(3^x)\, -\, 4\, =\, 0$

Using the factoring provided earlier, you should arrive at:

. . . . . $3^x\, =\, -\frac{4}{3}\, \mbox{ and }\, 3^x\, =\, \frac{1}{3}$

From what you've learned about exponential functions, you know that you cannot, by powers, turn a positive three into a negative value, so the first of the two equations above has "no solution".

Apply logs or the definition to solve the other exponential equation.

Note: The solution provided cannot be correct. If x = log_3(4/3), then log definitions and log rules give::

. . . . . $x\, +\, 1\, =\, \log_3(4/3)\, +\, \log_3(3)\, =\, \log_3(4)$

. . . . . $x\, +\, 2\, =\, \log_3(4/3)\, +\, \log_3(9)\, =\, \log_3(8)$

. . . . . $9^{x+1}\, =\, \left(3^{\log_3(4)}\right)^2\, =\, (4)^2\, =\, 16$

. . . . . $3^{x+2}\, =\, 3^{\log_3(8)}\, =\, 8$

. . . . . $9^{x+1}\, +\, 3^{x+2}\, -\, 4\, =\, 16\, +\, 8\, -\, 4\, =\, 20$

But the left-hand side is supposed to equal zero.

You might want to point out this typo to your instructor.
Memo to stapel:

Originally Posted by fabregas99
[snip]reall sorry on my first post just realised i put a + when it should have been a - imbetween 9^(x+1) and 3^(x+2).
[snip]