Math  /  Algebra

QuestionSolve for rr : log3(r+3)=1r=\begin{array}{l} \log _{3}(r+3)=-1 \\ r=\square \end{array}

Studdy Solution

STEP 1

1. The equation involves a logarithmic function with base 3.
2. We will use properties of logarithms to solve for r r .

STEP 2

1. Convert the logarithmic equation to its exponential form.
2. Solve the resulting equation for r r .

STEP 3

Convert the logarithmic equation to its exponential form. The equation log3(r+3)=1\log_{3}(r+3) = -1 can be rewritten using the definition of a logarithm:
If logb(a)=c\log_b(a) = c, then bc=ab^c = a.
Apply this to the given equation:
31=r+3 3^{-1} = r + 3

STEP 4

Calculate 313^{-1} and solve for r r :
31=13 3^{-1} = \frac{1}{3}
So, the equation becomes:
13=r+3 \frac{1}{3} = r + 3
Subtract 3 from both sides to solve for r r :
r=133 r = \frac{1}{3} - 3
Convert 3 to a fraction with a denominator of 3:
r=1393 r = \frac{1}{3} - \frac{9}{3}
r=193 r = \frac{1 - 9}{3}
r=83 r = \frac{-8}{3}
The value of r r is:
83 \boxed{-\frac{8}{3}}

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