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

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

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giorgiana1976's profile pic

Posted (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.

justaguide's profile pic

Posted (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

tonys538's profile pic

Posted (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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