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
- 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:

. . . . .

. . . . .

This makes the equation the same as:

. . . . .

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**:

. . . . .

Using the factoring provided earlier, you should arrive at:

. . . . .

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::

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

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