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

can anyone show me how to solve this.

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- Apr 24th 2009, 05:10 AMfabregas99exponential equation
9^(x+1) + 3^(x+2) - 4 = 0

can anyone show me how to solve this. - Apr 24th 2009, 05:28 AMmr fantastic
Since $\displaystyle 9^{x+1} = 9^x \cdot 9$ and $\displaystyle 9 = 3^2$ it follows that $\displaystyle 9^{x+1} = 9 \cdot (3^2)^x = 9 \cdot (3^x)^2$.

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

Therefore your equation can be re-written as

$\displaystyle 9 \cdot (3^x)^2 + 9 \cdot 3^x - 4 = 0 \Rightarrow (3 \cdot 3^x + 4) (3 \cdot 3^x - 1) = 0$. - Apr 24th 2009, 05:56 AMfabregas99thanks
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. - Apr 24th 2009, 06:12 AMmr fantastic
- Apr 24th 2009, 06:34 AMstapel
Use

**the rules for exponents**to restate in the manner outlined in a previous reply:

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

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

This makes the equation the same as:

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

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**:

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

Using the factoring provided earlier, you should arrive at:

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

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::

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

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

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

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

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

But the left-hand side is*supposed*to equal zero. (Surprised)

You might want to point out this typo to your instructor. (Wink) - Apr 24th 2009, 06:38 AMmr fantastic