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What is the solution of the exponential equation 9^(6 - x) - 8^x=0?

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greynose | Student, Undergraduate | (Level 2) Honors

Posted May 14, 2011 at 3:11 PM via web

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What is the solution of the exponential equation 9^(6 - x) - 8^x=0?

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted May 14, 2011 at 3:33 PM (Answer #1)

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We'll shift 8^x to the right side and we'll use logarithms to solve exponential equation.

We'll take the natural logarthim both sides:

ln 9^(6 - x) = ln 8^x

We'll apply the power rule for logarithms:

(6 - x)*ln 9 = x*ln 8

We'll re-write the equation:

6*ln 9 - x*ln 9 = x*ln 8

We'll add x*ln 9 both sides:

x*ln 9 + x*ln 8 = 6*ln 9

We'll factorize by x:

x*(ln 9 + ln 8) = ln 9^6

We'll divide by (ln 9 + ln 8):

x = ln 9^6/ln (8*9)

x = 13.1833/4.2766

The solution of the equation, rounded to four decimal places, is: x = 3.0826.

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted May 14, 2011 at 4:02 PM (Answer #2)

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The equation to be solved is 9^(6 - x) - 8^x = 0.

9^(6 - x) - 8^x = 0

=> 9^(6 - x) = 8^x

=> 9^6/9^x = 8^x

=> 9^6 = (9^x)(8^x)

=> 9^6 = 72^x

It is not possible to equate the base of both the sides, so we can use logarithm

6*log 9 = x*log 72

=> x = 6*log 9/log 72

=> x = 3.0826 ( approximately)

The solution of the equation is x = 3.0826

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tonys538 | TA , Undergraduate | (Level 1) Valedictorian

Posted October 31, 2014 at 2:33 PM (Answer #3)

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The equation 9^(6 - x) - 8^x=0 has to be solved for x. Here, it is only possible to find the solution as an expression involving logarithm as it is not possible to equate either the base or the exponent.

9^(6 - x) - 8^x=0

9^(6 - x) = 8^x

Take the log of both the sides

log(9^(6 - x)) = log(8^x)

Use the property log a^b = b*log a

(6 - x)*log 9 = x*log 8

Isolate x to one side

6*log 9 - x*log 9 = x*log 8

x*(log 8 + log 9) = 6*log 9

x*log 72 = 6*log 9

x = (6*log 9)/log 72

The solution of the equation 9^(6 - x) - 8^x=0 is x = (6*log 9)/log 72

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