Math  /  Algebra

QuestionSolve the following equation for xx. log3x+log37=1x=\begin{array}{l} \log _{3} x+\log _{3} 7=1 \\ x=\square \end{array}

Studdy Solution

STEP 1

What is this asking? We need to find the value of xx that satisfies the given logarithmic equation. Watch out! Remember the properties of logarithms, especially the one about adding logs with the same base!

STEP 2

1. Combine the logarithms
2. Rewrite the equation in exponential form
3. Isolate xx

STEP 3

When we add logarithms with the same base, we can multiply the arguments.
This is super useful!
So, log3x+log37\log_{3} x + \log_{3} 7 becomes log3(x7)\log_{3}(x \cdot 7).
Our equation now looks like this: log3(x7)=1 \log_{3}(x \cdot 7) = 1

STEP 4

A logarithm is just another way of writing an exponent.
The equation logba=c\log_{b} a = c means the same thing as bc=ab^c = a.
Here, our base is **3**, the exponent is **1**, and the result is x7x \cdot 7.

STEP 5

Let's rewrite our logarithmic equation log3(x7)=1\log_{3}(x \cdot 7) = 1 in exponential form: 31=x7 3^1 = x \cdot 7

STEP 6

We know that 313^1 is just **3**, so our equation becomes: 3=x7 3 = x \cdot 7

STEP 7

To get xx by itself, we'll divide both sides of the equation by **7**: 37=x77 \frac{3}{7} = \frac{x \cdot 7}{7}

STEP 8

The **7**s on the right-hand side divide to one, leaving us with: x=37 x = \frac{3}{7}

STEP 9

The value of xx that satisfies the equation is 37\frac{3}{7}.

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