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College Algebra & Trigonometry, 2018a

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168 CHAPTER 3. EXPONENTS AND LOGARITHMS<br />

We can also approximate the logarithms in the problem and solve for an approximate<br />

answer:<br />

(2x +3)ln5=ln17<br />

x ∗ 2 ln 5 + 3 ln 5 = ln 17<br />

3.2189x +4.8283 ≈ 2.8332<br />

−4.8283 ≈−4.8283<br />

3.2189x ≈−1.9951<br />

x ≈−0.620<br />

If you use the method of approximating, it’s important to make a good approximation.<br />

At least 4-5 decimal places are necessary for an accurate answer.<br />

Let’s look at an example that has variables on both sides of the equation:<br />

Example<br />

Solve for x.<br />

4 3x =9 2x−1<br />

We’ll use log base 10 in this problem.<br />

4 3x =9 2x−1<br />

log 4 3x =log9 2x−1<br />

3x ∗ log4=(2x − 1) log 9<br />

x ∗ 3log4=x ∗ 2log9− log 9

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