Solving a second-degree exponential equation with logarithms The following equation is given: 8^(2x)+8^x−20=0

Dangelo Rosario

Dangelo Rosario

Answered question

2022-09-09

Solving a second-degree exponential equation with logarithms
The following equation is given:
8 2 x + 8 x 20 = 0
The objective is to solve for x in terms of the natural logarithm ln.
I approach as follows:
log 8 ( 8 2 x ) = log 8 ( 8 x + 20 )
2 x = log 8 ( 8 x + 20 )
2 x = ln ( 8 x + 20 ) ln 8
x = ln ( 8 x + 20 ) 2 ln 8
and at this point I'm unable to proceed.

Answer & Explanation

Garrett Valenzuela

Garrett Valenzuela

Beginner2022-09-10Added 9 answers

Hint: Make the substitution u = 8 x . Now the equation becomes u 2 + u 20 = 0
Haiden Meyer

Haiden Meyer

Beginner2022-09-11Added 1 answers

We have
( 8 x ) 2 + 8 x 20 = 0
8 x = 1 ± 1 4 ( 20 ) 2 = 5 , 4
For real x , 8 x > 0
Then 8 x = 4 x ln 8 = ln 4
But ln 8 = ln ( 2 3 ) = 3 ln 2 and ln 4 = ln ( 2 2 ) = 2 ln 2

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