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# Solve...

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7(3^(x+1)) = 2 + 3 / 3^x

In the form a - log(b, 3) where a and b are natural numbers.

The answer is 1 - log(7,3)

Which I found by trial and error

Guest Nov 26, 2017
edited by Guest  Nov 26, 2017

#4
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$$7\cdot 3^{x+1} = 2+\dfrac{3}{3^{x}}\\ 21\cdot 3^{2x}-2\cdot 3^x - 3=0\\ \boxed{u=3^x}\\ 21u^2 - 2u -3 = 0\\ (3u+1)(7u-3)=0\\ u = \dfrac{-1}{3}(rej.) \quad OR\quad u = \dfrac{3}{7}\\ 3^{x-1} = \dfrac{1}{7}\\ (x-1) = \log_3\dfrac{1}{7}\\ x = 1 + \log_3(7^{-1})\\ x = 1 - \log_3(7)$$

MaxWong  Nov 27, 2017
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#1
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Solve for x :
7 3^(x + 1) = 3^(1 - x) + 2

Multiply both sides by 3^(x - 1):
7 3^(2 x) = 2 3^(x - 1) + 1

Subtract 2 3^(x - 1) + 1 from both sides:
-1 - 2 3^(x - 1) + 7 3^(2 x) = 0

The left hand side factors into a product with three terms:
1/3 (7 3^x - 3) (3^(x + 1) + 1) = 0

Multiply both sides by 3:
(7 3^x - 3) (3^(x + 1) + 1) = 0

Split into two equations:
7 3^x - 3 = 0 or 3^(x + 1) + 1 = 0

7 3^x = 3 or 3^(x + 1) + 1 = 0

Divide both sides by 7:
3^x = 3/7 or 3^(x + 1) + 1 = 0

Take the logarithm base 3 of both sides:
x = -(log(7/3))/(log(3)) or 3^(x + 1) + 1 = 0

Subtract 1 from both sides:
x = -(log(7/3))/(log(3)) or 3^(x + 1) = -1

3^(x + 1) = -1 has no solution since for all z element R, 3^z>0 and -1<0:
x = -(log(7/3))/(log(3))

Guest Nov 26, 2017
edited by Guest  Nov 26, 2017
#2
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I greatly appreciate your answer, but the answer must be in the form a - log(b, 3).

Where a and b are natural whole Numbers.

I’ve found that a = 1 and b = 7 by trial and error.

Guest Nov 26, 2017
#3
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This answer of: x = -(log(7/3))/(log(3))  is the same as:

1 - log(base3), 7 = -0.771243749, which the correct value of x, but not: 1 - log(base7), 3 =0.43542496, which is NOT correct value of x.

Guest Nov 26, 2017
#4
+6928
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$$7\cdot 3^{x+1} = 2+\dfrac{3}{3^{x}}\\ 21\cdot 3^{2x}-2\cdot 3^x - 3=0\\ \boxed{u=3^x}\\ 21u^2 - 2u -3 = 0\\ (3u+1)(7u-3)=0\\ u = \dfrac{-1}{3}(rej.) \quad OR\quad u = \dfrac{3}{7}\\ 3^{x-1} = \dfrac{1}{7}\\ (x-1) = \log_3\dfrac{1}{7}\\ x = 1 + \log_3(7^{-1})\\ x = 1 - \log_3(7)$$