1. Example 1 Solve Using Equal Powers Property Solve the equation. a. 4 9x 5 42 42 = – 4 x + 1 23x23x = b. Write original equation. SOLUTION a. 4 9x 5 42 42 = – 9x9x2 = – 5 Equal powers property 9x9x7 = Add 5 to each side. 9 7 x = Divide each side by 9.

2. Example 1 Solve Using Equal Powers Property ANSWER The solution is. Check this in the original equation. 9 7 b. 4 x 1 23x23x = + Write original equation. 23x23x = () x 12 + Rewrite 4 as 2 2 so powers have same base. 23x23x = 2 2 x 1 + () Power of a power property 3x3x = 2 () 1x + Equal powers property 3x3x = 2x2x 2 + Distributive property x = 2 Subtract 2x from each side.

3. Example 1 Solve Using Equal Powers Property ANSWER The solution is 2. Check this in the original equation.

4. Checkpoint Solve the equation. Solve Using Equal Powers Property 1. 23 23 = 2 7x 4 – ANSWER 1 1 4 6 2. = 3 x 2 –– 3 5x 6 3. = 5 x 3 – + 5 4x 9 4. = 2 x 2 – + 16 x 4

5. Checkpoint Solve Using Equal Powers Property 5. 36 8x 1 = 6 4x 1 – + ANSWER 4 1 5 Solve the equation. 6. = 10 x 3 –– 100 2x 9

6. Example 2 Take a Common Logarithm of Each Side Solve SOLUTION 3x3x = 5 Write original equation. 3x3x = 5.5. log 3 x = log 5 Take common logarithm of each side. x log 3 = log 5 Power property of logarithms x = Divide each side by log 3. log 5 log 3 x Use a calculator. 1.465 ≈

7. Example 2 Take a Common Logarithm of Each Side The solution is about 1.465. Check this in the original equation. ANSWER

8. Example 3 Take a Common Logarithm of Each Side Solve = 19. 10 3x 1 – SOLUTION Write original equation. = 1910 3x 1 – Take common logarithm of each side. = log 19log 10 3x 1 – = log 191 – 3x3x log 10 x x = = log 191 + Add 1 to each side. 3x3x x = Divide each side by 3. 3 log 191 + x Use a calculator. 0.760 ≈

9. Example 3 Take a Common Logarithm of Each Side CHECKYou can check the solution by substituting it into the original equation. Or, you can check the solution graphically by graphing each side of the original equation as a function. The two graphs intersect when x 0.760. ≈ 19 = y1y1 = 10 3x 1 – y2y2 and

10. Checkpoint Solve the equation. Take a Common Logarithm of Each Side 7. 2x2x = 9 ANSWER 3.170 8. 4x4x = 5 ANSWER 1.161 9. 3x3x = 40 ANSWER 3.358 ANSWER 0.233 10. = 510 3x 11. = 610 2x 5 + ANSWER 2.111 –

11. Checkpoint Solve the equation. Take a Common Logarithm of Each Side 12. = 1310 x ANSWER 1.230 – – 4 –

12. Example 4 Solve a Logarithmic Equation Solve. log 7 = () 4x4x3 – () x6 + SOLUTION Write original equation. log 7 = () 4x4x3 – () x6 + Equal logarithms property = 4x4x3 – x6 + Add 3 to each side. = x9 + 4x4x Subtract x from each side. = 93x3x Divide each side by 3. = 3x The solution is 3. Check this in the original equation. ANSWER

13. Example 5 Exponentiate Each Side Solve log 2 = () 3x3x1 + 4.4. SOLUTION log 2 = () 3x3x1 + 4 Write original equation. = log 2 () 3x3x1 + 2424 Exponentiate each side using base 2. 2 = 3x3x1 + 16 log b x b = x = 3x3x15 Subtract 1 from each side. = x5 Divide each side by 3. The solution is 5. Check this in the original equation. ANSWER

14. Checkpoint Solve the equation. Solve a Logarithmic Equation 13. ANSWER 7 2 2 16 = log 3 () 2x2x5 – () x2 + 15. = 2log 4 () 7x7x2 + 16. = 4log 3 () 5x5x1 + 14. log 5 () 8x8x9 – = () 3x3x1 +

15. Example 6 Check for Extraneous Solutions Solve Check for extraneous solutions. log = ) 3 – log 10x + ( x 2.2. SOLUTION Write original equation. log = ) 3 – log 10x + ( x2 Product property of logarithms 10x = ) 3 – log ( x2 [ ] Exponentiate each side using base 10. = 10 2 10x ) 3 – log ( x [ ] 10 10x = ) 3 – ( x100 10 log x = x 10x 2 = 30x – 100 Simplify.

16. Example 6 Check for Extraneous Solutions 10x 2 = 30x – 0 Subtract 100 from each side. 100 – = 0 Factor. 10 ) 5 – ( x ) 2 + ( x = Zero product property 5x = x – 2 or ANSWER The solution is 5. The solutions appear to be 5 and. However, when you check these in the original equation or use a graphic check as shown at the right, you can see that is the only solution. – 2 5x =

17. Example 7 Use Logarithms with an Exponential Model Radioactive Decay The exponential decay model for predicting the amount A of material left in a radioactive sample after t years is = 2 t/h AA0A0 – where A 0 is the initial amount of the substance and h is the half-life of the substance. Cesium is an element found in rocks and soil. A radioactive form of cesium, 137 Cs (read as “Cesium- 137 ”), has a half-life of about 30.2 years. How long does it take for 32 grams of 137 Cs to decay to 4 grams?

18. Example 7 Use Logarithms with an Exponential Model SOLUTION Write radioactive decay model. = 2 t/h AA0A0 – Substitute 4 for A, 32 for A 0, and 30.2 for h. = 2 t/30.2 432 – 8 1 Divide each side by 32. = 2 t/30.2 – log 2 2323 1 Take logarithm of each side using base 2. = 2 t/30.2 – log 2 = 3 – 30.2 t – log b b x x = Multiply each side by 30.2. = 90.6t –

19. Example 7 Use Logarithms with an Exponential Model It takes about 90.6 years for 32 grams of 137 Cs to decay to 4 grams. ANSWER