Log4^16+log1/2^(3x+1)=log1/4^(3x+1)

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First, let's use the property of logarithms that states log(a^b) = b * log(a) to simplify the left side of the equation:

log4^16 = 16 log(4
log1/2^(3x+1) = (3x+1) log(1/2)

So, the equation becomes:

16 log(4) + (3x+1) log(1/2) = log(1/4^(3x+1))

Next, we can simplify log(4) and log(1/2) using the change of base formula:

log(4) = log(2^2) = 2 * log(2
log(1/2) = log(2^-1) = -log(2)

Substitute these values into the equation:

16 2 log(2) + (3x+1) * (-log(2)) = log(1/4^(3x+1))

32 log(2) - (3x+1) log(2) = log(1/4^(3x+1))

Now, we can simplify the equation further by combining like terms:

(32 - 3x - 1) * log(2) = log(1/4^(3x+1))

(31 - 3x) * log(2) = log(1/4^(3x+1))

Finally, we can now convert this logarithmic equation to an exponential equation:

2^(31-3x) = 1/4^(3x+1)

2^(31-3x) = 1/(2^(2*(3x+1)))

2^(31-3x) = 2^(-6x - 2)

Now, we have two exponents with the same base, so we can set the exponents equal to each other:

31 - 3x = -6x - 2

31 = 3x - 6x - 2

31 = -3x - 2

33 = -3x

x = -11

So, the solution to the given logarithmic equation is x = -11.