Logarithm and Antilogarithm Test 20

$$\log_2(64)$$

$$(-1)^{19}+(-1)^{20}+(2)^5$$
$$=(-1)+(1)+2\times2\times2\times2\times2$$
$$=32$$

$$\textbf{Step 1 : The principle properties of logarithm:} \log_am^n = n\log_am \left( \textbf{Where m,a>0 and a}\neq 1\right)$$

Consider,

$${{\log }_{4}}\dfrac{{{t}^{2}}}{4}-2{{\log }_{4}}4{{t}^{4}}$$

Put $$t=-2$$ in above expression, We get

$$ ={{\log }_{4}}\dfrac{{{\left( -2 \right)}^{2}}}{4}-2{{\log }_{4}}4{{\left( -2 \right)}^{4}} $$

$$ ={{\log }_{4}}\left( \dfrac{4}{4}\right)-2{{\log }_{4}}\left(4\times 16\right) $$

$$ ={{\log }_{4}}1-2{{\log }_{4}}64 $$

$$ =0-2{{\log }_{4}}{{4}^{3}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( \because {{\log }_{a}}1=0 \right) $$

$$ =-2\times 3{{\log }_{4}}4\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( \because {{\log }_{a}}{{m}^{n}}=n{{\log }_{a}}m \right) $$

$$ =-6\,\,\,\,\,\,\,          \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( \therefore {{\log }_{a}}a=1 \right) $$

 

Hence,$$-6$$ is the answer.

$$\log_{2}{\cfrac{625}{10000}}$$

Mantissa of logarithm of $$719.3$$ to base $$10$$ is $$0.8569$$

$$2y=\log\left(12-5x-3x^2\right)$$

$$a=\log _{ 3 }{ 5 } =\cfrac { \log { 5 }  }{ \log { 3 }  } ,b=\cfrac { \log { 25 }  }{ \log { 7 }  } =\cfrac { \log { 5^2 }  }{ \log { 7 }  }=\cfrac { 2\log { 5 }  }{ \log { 7 }  } $$

$$\\(\frac{8^{-1}\cdot 5^3}{2^{-4}\cdot 625})\\= (\frac{2^4\cdot 5^3}{8\cdot 5^4})\\= (\frac{16}{8\cdot 5})\\=(\frac{2}{5})$$

$$Since, \ \log[(x+3)(x-3)]=log16\\x^2-3^2=16\\x^2=16+9\\=x^2=25\\\therefore\>x=\>\pm5\\but\>x\neq\>-5\>as\>log\>has\>domain\>of\>only\>+ve\>values\\\therefore\>x\>=\>5$$