# Find the inverse of y= log 1/4 ( 4*5^x) - 7

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y= log 1/4 ( 4*5^x) - 7

To find the enverse first we will add 7 to both sides:

==> y+ 7 = log 1/4 ( -4*5x)

Now we know from algorethim properties that:

log 1/4 (-4*5x) = log 1/4 4 + log 1/4 * 5x

= -1 + log 1/4 5x

==> y+ 7 = -1 + log 1/4 5x

Now add 1 to both sides:

==> y + 8 = log 1/4 5x

Now we will rewrite using the exponent:

==> 5x = (1/4)^(y+8)

==> x = (1/4)^(y+8) ]/ 5

Then the inverse for y s:

y^-1 =[ (1/4)^(x+8)]/ 5

The inverse of the function y= log 1/4 ( 4*5^x) - 7 is:

y + 7 = log 1/4 ( 4*5^x)

We'll use the product rule of logarithms:

log 1/4 ( 4*5^x) = log 1/4 (1*4) + log 1/4 (5^x)

We'll use the power rule for log 1/4 (5^x):

log 1/4 (5^x) = x*log 1/4 (5)

We'll re-write the function, isolating x to the right side:

y + 7 - log 1/4 (4) = x*log 1/4 (5)

But log 1/4 (4) = 1/log 4 (4^-1) = -1

We'll divide by log 1/4 (5):

x = (y + 7 + 1)/log 1/4 (5)

**f^-1(x) = (x + 8)/log 1/4 (5)**