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How do you solve #4-4 + log _ 9 (3x - 7) = 6#?

Answer 1

Liam Brennan

#x=(9^6+7)/3#

#color(red)(cancel(color(black)(4)))-color(red)(cancel(color(black)(4)))+log_9(3x-7)=6#

#=>log_9(3x-7)=6#

To undo the logarithm with base #9#, exponentiate each side.

#=>9^(log_9(3x-7))=9^6#

#=>3x-7=9^6=531441#

#=>3x=531448#

#=>x=531448/3=177149 1/3#

This can also be written as #x=(9^6+7)/3#.

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Answer 2

Lincoln Anderson

To solve (4 - 4 + \log_9(3x - 7) = 6):

Combine like terms on the left side of the equation.

[4 - 4 = 0]

[0 + \log_9(3x - 7) = 6]

Subtract 6 from both sides of the equation.

[\log_9(3x - 7) = 6 - 6]

[\log_9(3x - 7) = 0]

Rewrite the equation in exponential form.

[9^0 = 3x - 7]

Simplify (9^0), which equals 1.

[1 = 3x - 7]

Add 7 to both sides of the equation.

[1 + 7 = 3x - 7 + 7]

[8 = 3x]

Divide both sides by 3.

[x = \frac{8}{3}]

So, the solution to the equation is (x = \frac{8}{3}).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.