# What is the solution to: 1/16 = 64^(4x+7)I think you have to use natural log or log, but i'm not sure which.

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The equation `1/16 = 64^(4x+7)` has to be solved.

This can be done using logarithm of any base. Take logarithm to base 10 of both the sides.

`log(1/16) = log(64^(4x+7))`

`log 2^(-4) = log (2^6)^(4x+7)`

`log 2^(-4) = log 2^(6*(4x+7))`

Use the relation log `a^b = b*log a`

`-4*log 2 = 6*(4x+7)*log 2`

-4 = 24x + 42

24x = -46

x = `-23/12`

The solution of the equation `1/16 = 64^(4x+7)` is `x = -23/12`

We notice that we can create matching bases both sides. We'll write and 64 as powers of 2.

1/2^4 = (2^6)^(4x+7)

For the term from the left side, we'll apply the negative power rule:

1/2^4 = 2^-4

By definition, we'll multiply the exponents of the term from the right side:

2^-4 = 2^(24x + 42)

Since the bases are matching now, we can apply the one to one property of exponentials.

-4 = 24x + 42

We'll re-write the equation:

24x + 42 = -4

We'll isolate x to the left side:

24x = -4 - 42

24x = -46

We'll divide by 24:

x = -46/24

x = -23/12

**The solution of the equation is x = -23/12.**