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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)` ).
Then, apply the rule of logarithm which is `log_b a^n = n log_b a` .
`2x= 2/(log9) - 13`
(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` .
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`
`log9 (2x + 13 ) = 2`
Solve for x:
x = `(((2)/(log9)) - 13 )/2`
the value of x=(1/log^9)-13/2
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