(a)

\begin{align} 5^x & = 9 \\ \\ \text{Taking } \lg & \text{ of both sides,} \\ \lg 5^x & = \lg 9 \\ [\text{Power law}] \phantom{00000000} x\lg 5 & = \lg 9 \\ x & = {\lg 9 \over \lg 5} \\ & = 1.3652 \\ & \approx 1.37 \end{align}

(b)

\begin{align} 4e^{2x} & = 21 \\ e^{2x} & = {21 \over 4} \\ e^{2x} & = 5.25 \\ \\ \text{Taking } \ln & \text{ of both sides,} \\ \ln e^{2x} & = \ln 5.25 \\ [\text{Power law}] \phantom{00000000} 2x(\ln e) & = \ln 5.25 \\ [\ln e = \log_e e = 1] \phantom{000000000.} 2x(1) & = \ln 5.25 \\ 2x & = \ln 5.25 \\ x & = {\ln 5.25 \over 2} \\ & = 0.82911 \\ & \approx 0.829 \end{align}

(c)

\begin{align} 4 - 7^{2x} & = 1 \\ -7^{2x} & = -3 \\ 7^{2x} & = 3 \\ \\ \text{Taking } \lg & \text{ of both sides,} \\ \lg 7^{2x} & = \lg 3 \\ [\text{Power law}] \phantom{00000000} 2x \lg 7 & = \lg 3 \\ 2x & = {\lg 3 \over \lg 7} \\ x & = {\lg 3 \over \lg 7} \div 2 \\ & = 0.28228 \\ & \approx 0.282 \end{align}

(d)

\begin{align} 3^{x + 1} - 12 & = 0 \\ 3^{x + 1} & = 12 \\ \\ \text{Taking } \lg & \text{ of both sides,} \\ \lg 3^{x + 1} & = \lg 12 \\ [\text{Power law}] \phantom{00000000} (x + 1)\lg 3 & = \lg 12 \\ x + 1 & = {\lg 12 \over \lg 3} \\ x & = {\lg 12 \over \lg 3} - 1 \\ & = 1.2618 \\ & \approx 1.26 \end{align}

\begin{align} \text{When } & y = 12, \\ 12 & = 5e^{0.2x} \\ {12 \over 5} & = e^{0.2x} \\ \\ \text{Taking } \ln & \text{ of both sides,} \\ \ln {12 \over 5} & = \ln e^{0.2x} \\ \ln {12 \over 5} & = 0.2x (\ln e) \phantom{00000000} [\text{Power law}] \\ \ln {12 \over 5} & = 0.2x (1) \phantom{000000000.} [\ln e = \log_e e = 1] \\ \ln {12 \over 5} & = 0.2x \\ \\ \therefore x & = {\ln {12 \over 5} \over 0.2} \\ & = 4.3773 \\ & \approx 4.38 \end{align}

(i)

\begin{align} 4^x & = 9(5^x) \\ {4^x \over 5^x} & = 9 \\ \left[ {a^m \over b^m} = \left(a \over b\right)^m \right] \phantom{0000000000} \left(4 \over 5\right)^x & = 9 \end{align}

(ii)

\begin{align} 4^x & = 9(5^x) \\ \\ \text{Applying the } & \text{result from part (i),} \\ \left(4 \over 5\right)^x & = 9 \\ \\ \text{Taking } \lg & \text{ of both sides,} \\ \lg \left(4 \over 5\right)^x & = \lg 9 \\ [\text{Power law}] \phantom{00000000} x \lg {4 \over 5} & = \lg 9 \\ x & = {\lg 9 \over \lg {4 \over 5}} \\ & = -9.8466 \\ & \approx -9.85 \end{align}

(a)

\begin{align} 2^x . 3^x & = 10 \\ [ a^m \times b^m = (ab)^m ] \phantom{00000000} (2 \times 3)^x & = 10 \\ 6^x & = 10 \\ \\ \text{Taking } \lg & \text{ of both sides,} \\ \lg 6^x & = \lg 10 \\ \lg 6^x & = 1 \phantom{0000000000} [\lg 10 = \log_{10} 10 = 1] \\ [\text{Power law}] \phantom{0000000000} x \lg 6 & = 1 \\ x & = {1 \over \lg 6} \\ & = 1.2850 \\ & \approx 1.29 \end{align}

(b)

\begin{align} 2^{x + 1} & = 3^x \\ [ a^{m + n} = a^m \times a^n] \phantom{0000000000} 2^x \times 2^1 & = 3^x \\ 2(2^x) & = 3^x \\ 2 & = {3^x \over 2^x} \\ 2 & = \left(3 \over 2\right)^x \phantom{0000000000} \left[ {a^m \over b^m} = \left(a \over b\right)^m \right] \\ \\ \text{Taking } \lg & \text{ of both sides,} \\ \lg 2 & = \lg \left(3 \over 2\right)^x \\ \lg 2 & = x \lg \left(3 \over 2\right) \phantom{000000000} [\text{Power law}] \\ {\lg 2 \over \lg \left(3 \over 2\right)} & = x \\ \\ \therefore x & = 1.7095 \\ & \approx 1.71 \end{align}

(c)

\begin{align} 3^{x + 1}. 2^{x - 2} & = 21 \\ [ a^{m + n} = a^m \times a^n ] \phantom{00000000} (3^x \times 3^1). 2^{x - 2} & = 21 \\ 3(3^x)(2^{x - 2}) & = 21 \\ (3^x)(2^{x - 2}) & = {21 \over 3} \\ (3^x)(2^{x - 2}) & = 7 \\ \left[ a^{m - n} = {a^m \over a^n} \right] \phantom{0000000000} (3^x)\left(2^x \over 2^2\right) & = 7 \\ \left(3^x \over 1 \right)\left(2^x \over 4\right) & = 7 \\ {(3^x)(2^x) \over 4} & = 7 \\ (3^x)(2^x) & = 4(7) \\ (3^x)(2^x) & = 28 \\ [ a^m \times b^m = (ab)^m ] \phantom{0000000000000} (2 \times 3)^x & = 28 \\ 6^x & = 28 \\ \\ \text{Take } \lg & \text{ of both sides,} \\ \lg 6^x & = \lg 28 \\ [\text{Power law}] \phantom{000000000000000} x\lg 6 & = \lg 28 \\ x & = {\lg 28 \over \lg 6} \\ & = 1.8597 \\ & \approx 1.86 \end{align}

(i)

\begin{align} 9^x - 4 & = 3^{x + 1} \\ (3^2)^x - 4 & = 3^{x + 1} \\ [ (a^m)^n = a^{mn} ] \phantom{0000000000} 3^{2x} - 4 & = 3^{x + 1} \\ (3^x)^2 - 4 & = 3^{x + 1} \\ (3^x)^2 - 4 & = 3^x \times 3^1 \phantom{00000000} [ a^{m + n} = a^m \times a^n] \\ (3^x)^2 - 4 & = 3(3^x) \\ (3^x)^2 - 3(3^x) - 4 & = 0 \end{align}

(ii)

\begin{align} (3^x)^2 - 3(3^x) - 4 & = 0 \\ \\ \text{Let } & u = 3^x, \\ u^2 - 3u - 4 & = 0 \\ (u - 4)(u + 1) & = 0 \end{align} \begin{align} u - 4 & = 0 \phantom{00}&\text{or}\phantom{0000} u + 1 & = 0 \\ u & = 4 & u & = -1 \\ \\ \text{Since } & u = 3^x, \\ 3^x & = 4 & 3^x & = -1 \text{ (Reject, since } 3^x > 0) \\ \\ \text{Taking } & \lg \text{ of both sides,} \\ \lg 3^x & = \lg 4 \\ [\text{Power law}] \phantom{00000000} x\lg 3 & = \lg 4 \\ x & = {\lg 4 \over \lg 3} \\ & = 1.2618 \\ & \approx 1.26 \end{align}

(i)

\begin{align} 8^x + 11(2^x) & = 4^{x + {3 \over 2}} - 20 \\ 8^x + 11(2^x) & = 4^x \times 4^{3 \over 2} - 20 \phantom{00000000} [ a^{m + n} = a^m \times a^n] \\ 8^x + 11(2^x) & = 4^x \times 8 - 20 \\ 8^x + 11(2^x) & = 8(4^x) - 20 \\ (2^3)^x + 11(2^x) & = 8[(2^2)^x] - 20 \\ [ (a^m)^n = a^{mn} ] \phantom{00000000} 2^{3x} + 11(2^x) & = 8(2^{2x}) - 20 \\ (2^x)^3 + 11(2^x) & = 8(2^x)^2 - 20 \\ \\ \text{Let } & u = 2^x, \\ u^3 + 11u & = 8u^2 - 20 \\ \\ \therefore u^3 - 8u^2 & + 11u + 20 = 0 \end{align}

(ii)

\begin{align} \text{From } & \text{part (i),} \\ u^3 - 8u^2 & + 11u + 20 = 0 \\ \\ \text{Let } f(u) & = u^3 - 8u^2 + 11u + 20 \\ f(4) & = (4)^3 - 8(4)^2 + 11(4) + 20 \\ & = 0 \\ \\ \text{By Factor } & \text{theorem, } u - 4 \text{ is a factor.} \end{align}
$$ \require{enclose} \begin{array}{rll} u^2 \phantom{0} - 4u \phantom{/} - 5\phantom{0000000}\\ u - 4 \enclose{longdiv}{u^3 - 8u^2 + 11u + 20\phantom{0}}\kern-.2ex \\ -\underline{(u^3 - 4u^2){\phantom{0000000000}}} \\ -4u^2 + 11u + 20\phantom{0} \\ -\underline{(-4u^2 + 16u){\phantom{0000/}}} \\ - 5u + 20\phantom{0} \\ -\underline{(-5u + 20)} \\ 0\phantom{0} \end{array} $$
\begin{align} \text{Polynomial} & = \text{Divisor} \times \text{Quotient} + \text{Remainder} \\ f(u) & = u^3 - 8u^2 + 11u + 20 \\ & = (u - 4)(u^2 - 4u - 5) \\ & = (u - 4)(u - 5)(u + 1) \\ \\ \therefore 0 & = u^3 - 8u^2 + 11u + 20 \\ 0 & = (u - 4)(u - 5)(u + 1) \end{align} \begin{align} u - 4 & = 0 \phantom{00}&\text{or}\phantom{0000} u - 5 & = 0 \phantom{00}&\text{or}\phantom{0000} u + 1 & = 0 \\ u & = 4 & u & = 5 & u & = -1 \\ \\ \text{Since } & u = 2^x, & \text{Since } & u = 2^x, & \text{Since } & u = 2^x, \\ 2^x & = 4 & 2^x & = 5 & 2^x & = -1 \text{ (Reject, since } 2^x > 0) \\ 2^x & = 2^2 & \lg 2^x & = \lg 5 \\ & & x\lg 2 & = \lg 5 \\ \therefore x & = 2 & x & = {\lg 5 \over \lg 2} \\ & & & = 2.3219 \\ & & & \approx 2.32 \end{align}

(a)

\begin{align} 2e^{2x} - 3e^x & = 2 \\ [ a^{mn} = (a^m)^n] \phantom{00000000} 2(e^x)^2 - 3e^x & = 2 \\ \\ \text{Let } & y = e^x, \\ 2y^2 - 3y & = 2 \\ 2y^2 - 3y - 2 & = 0 \\ (2y + 1)(y - 2) & = 0 \end{align} \begin{align} y - 2 & = 0 \phantom{00}&\text{or}\phantom{0000} 2y + 1 & = 0 \\ y & = 2 & 2y & = -1 \\ & & y & = -{1 \over 2} \\ \\ \text{Since } & y = e^x, & \text{Since } & y = e^x, \\ e^x & = 2 & e^x & = -{1 \over 2} \text{ (Reject, since } e^x > 0) \\ \\ \text{Take } \ln & \text{ of both sides,} \\ \ln e^x & = \ln 2 \\ [\text{Power law}] \phantom{00000000} x\ln e & = \ln 2 \\ x (1) & = \ln 2 \\ x & = 0.69314 \\ & \approx 0.693 \end{align}

(b)

\begin{align} e^x & = 7 - 12e^{-x} \\ e^x & = 7 - 12\left(1 \over e^x\right) \\ e^x & = 7 - {12 \over e^x} \\ \\ \text{Let } & y = e^x, \\ y & = 7 - {12 \over y} \\ y^2 & = 7y - 12 \\ y^2 - 7y + 12 & = 0 \\ (y - 4)(y - 3) & = 0 \end{align} \begin{align} y - 4 & = 0 \phantom{00}&\text{or}\phantom{0000} y - 3 & = 0 \\ y & = 4 & y & = 3 \\ \\ \text{Since } & y = e^x, & \text{Since } & y = e^x, \\ e^x & = 4 & e^x & = 3 \\ \\ \text{Taking } & \ln \text{ of both sides,} & \text{Taking } & \ln \text{ of both sides,} \\ \ln e^x & = \ln 4 & \ln e^x & = \ln 3 \\ x(\ln e) & = \ln 4 & x(\ln e) & = \ln 3 \\ x (1) & = \ln 4 & x(1) & = \ln 3 \\ x & = \ln 4 & x & = \ln 3 \\ & = 1.3862 & & = 1.0986 \\ & \approx 1.39 & & \approx 1.10 \end{align}

(c)

\begin{align} e^{3x} + 2e^x & = 3e^2x \\ (e^x)^3 + 2e^x & = 3(e^x)^2 \phantom{0000000000} [ a^{mn} = (a^m)^n ] \\ \\ \text{Let } & y = e^x, \\ y^3 + 2y & = 3y^2 \\ y^3 - 3y^2 + 2y & = 0 \\ y(y^2 - 3y + 2) & = 0 \\ y(y - 1)(y - 2) & = 0 \end{align} \begin{align} y & = 0 \phantom{00}&\text{or}\phantom{0000} y - 1 & = 0 \phantom{00}&\text{or}\phantom{0000} y - 2 & = 0 \\ & & y & = 1 & y & = 2 \\ \\ \text{Since } & y = e^x, & \text{Since } & y = e^x, & \text{Since } & y = e^x, \\ e^x & = 0 \text{ (Reject)} & e^x & = 1 & e^x & = 2 \\ \\ & & \text{Take } \ln & \text{ of both sides,} & \text{Take } \ln & \text{ of both sides,} \\ & & \ln e^x & = \ln 1 & \ln e^x & = \ln 2 \\ & & x\ln e & = 0 & x\ln e & = 0.69314 \\ & & x (1) & = 0 & x(1) & = 0.69314 \\ & & x & = 0 & x & = 0.69314 \\ & & & & & \approx 0.693 \end{align}

(d)

\begin{align} 2e^x & = 7\sqrt{e^x} - 3 \\ \\ \text{Let } & y = e^x, \\ 2y & = 7\sqrt{y} - 3 \\ 2(\sqrt{y})^2 & = 7\sqrt{y} - 3 \\ 2(\sqrt{y})^2 - 7\sqrt{y} + 3 & = 0 \\ (2\sqrt{y} - 1)(\sqrt{y} - 3) & = 0 \end{align} \begin{align} 2\sqrt{y} - 1 & = 0 \phantom{00}&\text{or}\phantom{0000} \sqrt{y} - 3 & = 0 \\ 2\sqrt{y} & = 1 & \sqrt{y} & = 3 \\ \sqrt{y} & = 0.5 & y & = 3^2 \\ y & = (0.5)^2 & y & = 9 \\ y & = 0.25 \\ \\ \text{Since } & y = e^x, & \text{Since } & y = e^x, \\ e^{x} & = 0.25 & e^{x} & = 9 \\ \\ \text{Take } \ln & \text{ of both sides,} & \text{Take } \ln & \text{ of both sides,} \\ \ln e^{x} & = \ln 0.25 & \ln e^x & = \ln 9 \\ x(\ln e) & = \ln 0.25 & x (\ln e) & = \ln 9 \\ x(1) & = \ln 0.25 & x(1) & = \ln 9 \\ x & = \ln 0.25 & x & = \ln 9 \\ & = -1.3862 & & = 2.1972 \\ & \approx -1.39 & & \approx 2.20 \end{align}

(a)

\begin{align} x^3 & = e^{6x - 1} \\ \\ \text{Take } \ln & \text{ of both sides,} \\ \ln x^3 & = \ln e^{6x - 1} \\ 3\ln x & = (6x - 1)(\ln e) \phantom{00000000000} [\text{Power law}] \\ 3\ln x & = (6x - 1)(1) \\ 3\ln x & = 6x - 1 \\ \ln x & = 2x - {1 \over 3} \\ \\ \therefore a = 2 &, b = -{1 \over 3} \end{align}

(b)

\begin{align} xe^{-x} & = 2.46 \\ \\ \text{Take } \ln & \text{ of both sides,} \\ \ln (xe^{-x}) & = \ln 2.46 \\ [\text{Product law}] \phantom{00000000000} \ln x + \ln e^{-x} & = \ln 2.46 \\ [\text{Power law}] \phantom{00000000} \ln x + (-x)(\ln e) & = \ln 2.46 \\ \ln x + (-x)(1) & = \ln 2.46 \\ \ln x - x & = \ln 2.46 \\ \ln x & = x + \ln 2.46 \\ \ln x & = x + 0.90016 \\ \ln x & = x + 0.9 \\ \\ \therefore a = 1, b & = 0.9 \end{align}

(c)

\begin{align} (xe^x)^2 & = 30e^{-x} \\ \\ \text{Take } \ln & \text{ of both sides,} \\ \ln (xe^x)^2 & = \ln (30e^{-x}) \\ [\text{Power law}] \phantom{00000000} 2\ln (xe^x) & = \ln (30e^{-x}) \\ 2(\ln x + \ln e^x) & = \ln 30 + \ln e^{-x} \phantom{00000000} [\text{Product law}] \\ 2\ln x + 2\ln e^x & = \ln 30 + \ln e^{-x} \\ 2\ln x + 2x \ln e & = \ln 30 + (-x)\ln e \\ 2\ln x + 2x (1) & = \ln 30 + (-x)(1) \\ 2\ln x + 2x & = \ln 30 - x \\ 2\ln x & = - x - 2x + \ln 30 \\ 2\ln x & = -3x + \ln 30 \\ \ln x & = -{3 \over 2}x + {\ln 30 \over 2} \\ \ln x & = -{3 \over 2}x + 1.70 \\ \\ \therefore a = -{3 \over 2}, & \phantom{0} b = 1.70 \end{align}

(a)

\begin{align} 5^{x - 1} \times 3^{x + 2} & = 10 \\ \left[ a^{m - n} = {a^m \over a^n} \right] \phantom{0000000000} {5^x \over 5^1} \times 3^{x + 2} & = 10 \\ {5^x \over 5} \times 3^{x + 2} & = 10 \\ 5^x \times 3^{x + 2} & = 5(10) \\ 5^x \times 3^{x + 2} & = 50 \\ [ a^{m + n} = a^m \times a^n] \phantom{00000000} 5^x \times 3^x \times 3^2 & = 50 \\ 5^x \times 3^x \times 9 & = 50 \\ 5^x \times 3^x & = {50 \over 9} \\ [ a^m \times b^m = (ab)^m ] \phantom{00000000000.} (5 \times 3)^x & = {50 \over 9} \\ 15^x & = {50 \over 9} \\ \\ \text{Take } \lg & \text{ of both sides,} \\ \lg 15^x & = {50 \over 9} \\ [\text{Power law}] \phantom{0000000000000} x \lg 15 & = {50 \over 9} \\ x & = {{50 \over 9} \over \lg 15} \\ & = 0.63322 \\ & \approx 0.633 \end{align}

(b)

\begin{align} 2^{2x} \times 5^{x + 1} & = 7 \\ [ a^{mn} = (a^m)^n] \phantom{00000000.} (2^2)^x \times 5^{x + 1} & = 7 \\ 4^x \times 5^{x + 1} & = 7 \\ [ a^{m + n} = a^m \times a^n] \phantom{00000000} 4^x \times 5^x \times 5^1 & = 7 \\ 4^x \times 5^x \times 5 & = 7 \\ 4^x \times 5^x & = {7 \over 5} \\ [ a^m \times b^m = (a \times b)^m ] \phantom{00000000000.} (4 \times 5)^x & = {7 \over 5} \\ 20^x & = {7 \over 5} \\ \\ \text{Taking } \lg & \text{ on both sides,} \\ \lg 20^x & = \lg {7 \over 5} \\ [\text{Power law}] \phantom{000000000000.} x \lg 20 & = \lg {7 \over 5} \\ x & = {\lg {7 \over 5} \over \lg 20} \\ & = 0.11231 \\ & \approx 0.112 \end{align}

(c)

\begin{align} 4(3^{2x}) & = e^x \\ [ a^{mn} = (a^m)^n] \phantom{00000000} 4[ (3^2)^x] & = e^x \\ 4(9^x) & = e^x \\ 4 & = {e^x \over 9^x} \\ 4 & = \left(e \over 9\right)^x \phantom{0000000000} \left[ {a^m \over b^m} = \left(a \over b\right)^m \right] \\ \\ \text{Taking } \lg & \text{ of both sides,} \\ \lg 4 & = \lg \left(e \over 9\right)^x \\ \lg 4 & = x \lg \left(e \over 9\right) \phantom{000000000} [\text{Power law}] \\ {\lg 4 \over \lg \left(e \over 9\right)} & = x \\ −1.1579 & = x \\ -1.16 & \approx x \end{align}

(d)

\begin{align} 3^x \times 10^{2x} & = 4 \times 20^{x - 2} \\ [ a^{mn} = (a^m)^n] \phantom{00000000} 3^x \times (10^2)^x & = 4 \times 20^{x - 2} \\ 3^x \times 100^x & = 4 \times 20^{x - 2} \\ [ a^m \times b^m = (ab)^m ] \phantom{000000000} (3 \times 100)^x & = 4 \times 20^{x - 2} \\ 300^x & = 4 \times 20^{x - 2} \\ 300^x & = 4 \times {20^x \over 20^2} \phantom{00000000} \left[ a^{m - n} = {a^m \over a^n} \right] \\ 300^x & = {4 \over 1} \times {20^x \over 400} \\ 300^x & = {20^x \over 100} \\ {300^x \over 20^x} & = {1 \over 100} \\ \left[ {a^m \over b^m} = \left(a \over b\right)^m \right] \phantom{000000000} \left(300 \over 20\right)^x & = {1 \over 100} \\ 15^x & = {1 \over 100} \\ \\ \text{Take } \lg & \text{ of both sides,} \\ \lg 15^x & = \lg {1 \over 100} \\ [\text{Power law}] \phantom{000000000000} x\lg 15 & = \lg {1 \over 100} \\ x & = { \lg {1 \over 100} \over \lg 15} \\ & = −1.7005 \\ & \approx -1.70 \end{align}

\begin{align} 4^{x + 3} & = 32(2^{x + y}) \\ (2^2)^{x + 3} & = 32(2^{x + y}) \\ [ (a^m)^n = a^{mn} ] \phantom{0000000000} 2^{2(x + 3)} & = 32(2^{x + y}) \\ 2^{2x + 6} & = 32(2^{x + y}) \\ 2^{2x + 6} & = 2^5 (2^{x + y}) \\ 2^{2x + 6} & = 2^{5 + (x + y)} \phantom{0000000000} [a^m \times a^n = a^{m + n} ] \\ 2^{2x + 6} & = 2^{5 + x + y} \\ \\ \therefore 2x + 6 & = 5 + x + y \\ 2x - x & = 5 - 6 + y \\ x & = y - 1\phantom{000} \text{ --- (1)} \\ \\ \\ \text{Substitute (1) into } & 9^x + 3^y = 10, \\ 9^{y - 1} + 3^y & = 10 \\ \left[ a^{m - n} = {a^m \over a^n} \right] \phantom{0000000000} {9^y \over 9^1} + 3^y & = 10 \\ {9^y \over 9} + 3^y & = 10 \\ 9^y + 9(3^y) & = 90 \\ (3^2)^y + 9(3^y) & = 90 \\ [ (a^m)^n = a^{mn} ] \phantom{00000000} 3^{2y} + 9(3^y) & = 90 \\ (3^y)^2 + 9(3^y) & = 90 \\ \\ \text{Let } & u = 3^y, \\ u^2 + 9u & = 90 \\ u^2 + 9u - 90 & = 0 \\ (u - 6)(u + 15) & = 0 \end{align} \begin{align} u - 6 & = 0 \phantom{00}&\text{or} \phantom{0000} u + 15 & = 0 \\ u & = 6 & u & = -15 \\ \\ \text{Since } & u = 3^y, & \text{Since } & u = 3^y, \\ 3^y & = 6 & 3^y & = -15 \text{ (Reject, since } 3^y > 0) \\ \\ \text{Take } \lg & \text{ of both sides,} \\ \lg 3^y & = \lg 6 \\ [\text{Power law}] \phantom{00000000} y \lg 3 & = \lg 6 \\ y & = {\lg 6 \over \lg 3} \\ \\ \text{Substitute } & y = {\lg 6 \over \lg 3} \text{ into (1),} \\ x & = {\lg 6 \over \lg 3} - 1 \end{align}
$$ \therefore x = {\lg 6 \over \lg 3} - 1 \approx 0.631, \phantom{0} y = {\lg 6 \over \lg 3} \approx 1.63 $$