Properties of Logarithms and Exponents

Sol A Cubing both sides of $x^{2/3}=4$ gives $x^2=4^3=64$, and then taking the square root of both sides gives $x=\pm8$.

Sol B

$$\ln\left(\frac{x^4}{(2x-1)^3(7x-5)^8}\right)$$

$$=\ln(x^4)-\ln[(2x-1)^3(7x-5)^8]=\ln(x^4)-[\ln((2x-1)^3)+\ln((7x-5)^8)]$$

$$=4\ln x-[3\ln(2x-1)+8\ln(7x-5)]= 4\ln x-3\ln(2x-1)-8\ln(7x-5)$$

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Sol C $2\ln(3x-4)-5\ln(2x-7)=\ln((3x-4)^2)-\ln((2x-7)^5)$ $=\ln\left(\frac{(3x-4)^2}{(2x-7)^5}\right)$.

Sol 1

$$\frac{x^3 (x^4)^5}{x^7 (x^2)^4}=\frac{x^3 x^{20}}{x^7 x^8}$$

$$=\frac{x^{23}}{x^{15}}=x^8$$

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Sol 2

$$\ln(\frac{(x^2+4)^5 \sqrt[3]{4x-3}}{\sqrt{3x-5} (7x+2)^9})$$

$$=\ln(\frac{(x^2+4)^5 (4x-3)^{1/3}}{(3x-5)^{1/2}(7x+2)^9})$$

$$=\ln((x^2+4)^5 (4x-3)^{1/3})-\ln((3x-5)^{1/2} (7x+2)^9)$$

$$=\ln((x^2+4)^5)+\ln((4x-3)^{1/3})-(\ln((3x-5)^{1/2})+\ln((7x+2)^9))$$

$$=5\ln(x^2+4)+1/3\ln(4x-3)-(1/2\ln(3x-5)+9\ln(7x+2))$$

$$=5\ln(x^2+4)+1/3\ln(4x-3)-1/2\ln(3x-5)-9\ln(7x+2)$$

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Sol 3 $\log_b (\frac {b^5 x^2}{y^3})=\log_b(b^5 x^2)-\log_b(y^3)$ $=\log_b b^5+\log_b x^2-\log_b y^3=5+2\log_b x-3\log_b y$ $=5+2(2.3)-3(3.1)=0.3$.

Sol 4

$$1/2\log_5(4x-3)-1/2\log_5(3x+1)-3/2\log_5(x^2+6)$$

$$=1/2(\log_5(4x-3)-\log_5(3x+1)-3\log_5(x^2+6))$$

$$=1/2(\log_5(4x-3)-(\log_5(3x+1)+3\log_5(x^2+6)))$$

$$=1/2(\log_5(4x-3)-(\log_5(3x+1)+\log_5(x^2+6)^3))$$

$$=1/2(\log_5(4x-3)-\log_5((3x+1)(x^2+6)^3))$$

$$=1/2\log_5\frac{4x-3}{(3x+1)(x^2+6)^3}$$

$$=\log_5\sqrt{\frac{4x-3}{(3x+1)(x^2+6)^3}}$$

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Sol 5 $(\log_9 3)(\log_5 1/25)=(\log_9 \sqrt{9})(\log_5 25^{-1})$ $=(\log_9 9^{1/2})(-\log_5 25)=(1/2)(-2)=-1$.

Sol 6 $\ln(\ln e)+\log_2 8=\ln 1+\log_2 2^3=0+3=3$.

Sol 7 $\log_4 \frac{36}{5}+\log_4 \frac{10}{9}$ $=\log_4(36/5)(10/9)=\log_4 8=3/2$ since $4^{3/2}=(\sqrt{4})^3=2^3=8$.

Sol 8 Since $\ln x =r$ and $\ln b=s$, $e^r=x$ and $e^s=b$. Then $b^{r/s}=(e^s)^{r/s}=e^r=x$, so $\log_b x =r/s$.

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