# Solve the logarithmic expression `log 9 ^(x+8) + log 9 ^(x+5)=2` a. exact solution b. what is the decimal approximation ( rounded 3 decimals)I am not sure if I wrote it out right ?

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`log 9^(x+8) + log 9^(x+5) = 2`

Apply the product rule of logarithm which is `log_b a + log_b c = log_b(a*c)`.

`log (9^(x+8) * 9^(x+5) )=2`

To multiply same base, add the exponents (`a^m*a^n=a^(m+n)` ).

`log 9^(x+8+x+5)=2`

`log 9^(2x+13)=2`

Then, apply the rule of logarithm which is `log_b a^n = n log_b a` .

`(2x+13)log9=2`

`2x+13=2/(log9)`

`2x= 2/(log9) - 13`

`x= (2/(log9)-13)/2`

`x= 1/(log9)-13/2`

**(a) The exact solution to the given equation is `x=1/(log9)-13/2` .**

**(b) When rounded to three decimal places, the approximate value of the solution is `x=-6.045` .**

**Sources:**

The property of logs allows us to just drop the exponent down to get:

`(x+8)log9 + (x+5)log 9 = 2`

Pull out the log9 term and get :

l`og9 ( (x + 8) + (x + 5) ) = 2`

Simplify :

`log9 (2x + 13 ) = 2`

Solve for x:

x = `(((2)/(log9)) - 13 )/2`

the value of x=(1/log^9)-13/2