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Solutions to Try Its

1. a. log10(1,000,000)=6{\mathrm{log}}_{10}\left(1,000,000\right)=6 is equivalent to 106=1,000,000{10}^{6}=1,000,000 b. log5(25)=2{\mathrm{log}}_{5}\left(25\right)=2 is equivalent to 52=25{5}^{2}=25 2. a. 32=9{3}^{2}=9 is equivalent to log3(9)=2{\mathrm{log}}_{3}\left(9\right)=2 b. 53=125{5}^{3}=125 is equivalent to log5(125)=3{\mathrm{log}}_{5}\left(125\right)=3 c. 21=12{2}^{-1}=\frac{1}{2} is equivalent to log2(12)=1{\text{log}}_{2}\left(\frac{1}{2}\right)=-1 3. log121(11)=12{\mathrm{log}}_{121}\left(11\right)=\frac{1}{2} (recalling that 121=(121)12=11\sqrt{121}={\left(121\right)}^{\frac{1}{2}}=11 ) 4. log2(132)=5{\mathrm{log}}_{2}\left(\frac{1}{32}\right)=-5 5. It is not possible to take the logarithm of a negative number in the set of real numbers. 6. It is not possible to take the logarithm of a negative number in the set of real numbers.

Solutions to Odd-Numbered Exercises

1. A logarithm is an exponent. Specifically, it is the exponent to which a base b is raised to produce a given value. In the expressions given, the base b has the same value. The exponent, y, in the expression by{b}^{y} can also be written as the logarithm, logbx{\mathrm{log}}_{b}x, and the value of x is the result of raising b to the power of y. 3. Since the equation of a logarithm is equivalent to an exponential equation, the logarithm can be converted to the exponential equation by=x{b}^{y}=x, and then properties of exponents can be applied to solve for x. 5. The natural logarithm is a special case of the logarithm with base b in that the natural log always has base e. Rather than notating the natural logarithm as loge(x){\mathrm{log}}_{e}\left(x\right), the notation used is ln(x)\mathrm{ln}\left(x\right). 7. ac=b{a}^{c}=b 9. xy=64{x}^{y}=64 11. 15b=a{15}^{b}=a 13. 13a=142{13}^{a}=142 15. en=w{e}^{n}=w 17. logc(k)=d{\text{log}}_{c}\left(k\right)=d 19. log19y=x{\mathrm{log}}_{19}y=x 21. logn(103)=4{\mathrm{log}}_{n}\left(103\right)=4 23. logy(39100)=x{\mathrm{log}}_{y}\left(\frac{39}{100}\right)=x 25. ln(h)=k\text{ln}\left(h\right)=k 27. x=23=18x={2}^{-3}=\frac{1}{8} 29. x=33=27x={3}^{3}=27 31. x=912=3x={9}^{\frac{1}{2}}=3 33. x=63=1216x={6}^{-3}=\frac{1}{216} 35. x=e2x={e}^{2} 37. 32 39. 1.06 41. 14.125 43. 12\frac{1}{2} 45. 4 47. –3 49. –12 51. 0 53. 10 55. 2.708 57. 0.151 59. No, the function has no defined value for = 0. To verify, suppose = 0 is in the domain of the function f(x)=log(x)f\left(x\right)=\mathrm{log}\left(x\right). Then there is some number n such that n=log(0)n=\mathrm{log}\left(0\right). Rewriting as an exponential equation gives: 10n=0{10}^{n}=0, which is impossible since no such real number n exists. Therefore, = 0 is not the domain of the function f(x)=log(x)f\left(x\right)=\mathrm{log}\left(x\right). 61. Yes. Suppose there exists a real number x such that lnx=2\mathrm{ln}x=2. Rewriting as an exponential equation gives x=e2x={e}^{2}, which is a real number. To verify, let x=e2x={e}^{2}. Then, by definition, ln(x)=ln(e2)=2\mathrm{ln}\left(x\right)=\mathrm{ln}\left({e}^{2}\right)=2. 63. No; ln(1)=0\mathrm{ln}\left(1\right)=0, so ln(e1.725)ln(1)\frac{\mathrm{ln}\left({e}^{1.725}\right)}{\mathrm{ln}\left(1\right)} is undefined. 65. 2

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  • Precalculus. Provided by: OpenStax Authored by: Jay Abramson, et al.. Located at: https://openstax.org/books/precalculus/pages/1-introduction-to-functions. License: CC BY: Attribution. License terms: Download For Free at : http://cnx.org/contents/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175..