# How do you find the inverse of #g(x)= log_8(4 x+6)#?

To find the inverse of ( g(x) = \log_8(4x + 6) ), follow these steps:

- Replace ( g(x) ) with ( y ): ( y = \log_8(4x + 6) ).
- Swap the roles of ( x ) and ( y ), and solve for ( y ).
- Rewrite the equation with ( x ) as the subject.

[ x = \log_8(4y + 6) ] [ 8^x = 4y + 6 ] [ 4y = 8^x - 6 ] [ y = \frac{8^x - 6}{4} ]

So, the inverse of ( g(x) = \log_8(4x + 6) ) is ( g^{-1}(x) = \frac{8^x - 6}{4} ).

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I tried this:

I would first use the definition of log to write:

hope it helps!

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To find the inverse of ( g(x) = \log_8(4x + 6) ), switch the roles of ( x ) and ( y ), then solve for ( y ).

So, [ x = \log_8(4y + 6) ] [ 8^x = 4y + 6 ] [ 4y = 8^x - 6 ] [ y = \frac{8^x - 6}{4} ]

Therefore, the inverse function of ( g(x) ) is ( g^{-1}(x) = \frac{8^x - 6}{4} ).

By signing up, you agree to our Terms of Service and Privacy Policy

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.

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.

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.

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