Respuesta :
1.log2 32=5
lne=1
lg100=2
5+1-2=4
2. 4log9 x= 2log3 x=log3 x^2
log3 x*x^2=9
log3 x^3=9
log 3 x=3
x=27
lne=1
lg100=2
5+1-2=4
2. 4log9 x= 2log3 x=log3 x^2
log3 x*x^2=9
log3 x^3=9
log 3 x=3
x=27
To simplify we need recall and apply the properties and laws of logarithms.
1)
[tex] log_{2}(32) + ln(e) + log_{10}(100) [/tex]
[tex] log_{2}(32) + log_{e}(e) + log_{10}(100)[/tex]
We need consider the base of each logarithm and express the numbers in the parentheses to each base raised to a certain index. or exponent.
That is
[tex] {2}^{5} = 32 \\ \\ {10}^{2} = 100[/tex]
As for the middle expression, the base and the number are equal so let us keep it for now.
Our expression now becomes,
[tex] log_{2}( {2}^{5} ) + log_{e}(e) + log_{10}( {10}^{2} )[/tex]
Recall this law of logarithm,
[tex] log_{a}( {m}^{n} ) = n log_{a}(m)[/tex]
[tex]5 log_{2}( {2}) + log_{e}(e) + 2log_{10}( {10} )[/tex]
Recall again that,
[tex] log_{a}(a) ,a \ne0 \: or \: 1[/tex]
Therefore our expression becomes,
[tex]5 (1) + (1) + 2(1) = 5 + 1 + 2 = 8[/tex]
2) We use change of base to solve this logarithmic equation.
[tex] log_{3}(x) + 4log_{9}(x) = 9[/tex]
[tex]log_{3}(x) + log_{9}( {x}^{4} ) = 9[/tex]
It will be easier and faster to change the base to 3.
Recall that,
[tex]\log_{x}(y)=\frac{log_{a}(y)}{log_{a}(x)}[/tex]
We apply this law to obtain,
[tex]log_{3}(x) + \frac{log_{3}( {x}^{4})}{log_{3}(9) } = 9[/tex]
[tex]log_{3}(x) + \frac{log_{3}( {x}^{4})}{log_{3}( {3}^{2} ) } = 9[/tex]
[tex]log_{3}(x) + \frac{log_{3}( {x}^{4})}{2log_{3}( {3} ) } = 9[/tex]
[tex]log_{3}(x) + \frac{log_{3}( {x}^{4})}{2(1) } = 9[/tex]
[tex]log_{3}(x) + \frac{log_{3}( {x}^{4})}{2} = 9[/tex]
multiplying through by 2 gives,
[tex]2log_{3}(x) + log_{3}( {x}^{4} ) = 18[/tex]
[tex]log_{3}( {x}^{2} ) + log_{3}( {x}^{4} ) = 18[/tex]
[tex] log_{3}( {x}^{4} \times {x}^{2} ) = 18[/tex]
We apply the multiplication law of exponents to obtain,
[tex] log_{3}( {x}^{4 + 2} ) = 18[/tex]
[tex] log_{3}( {x}^{6} ) = 18[/tex]
We take antilogarithms to get,
[tex] {x}^{6} = {3}^{18} [/tex]
[tex] {x}^{6} = ( {3}^{3} ) ^{6} [/tex]
[tex]x = {3}^{3} [/tex]
[tex]x = 27[/tex]
Hence x is 27.
1)
[tex] log_{2}(32) + ln(e) + log_{10}(100) [/tex]
[tex] log_{2}(32) + log_{e}(e) + log_{10}(100)[/tex]
We need consider the base of each logarithm and express the numbers in the parentheses to each base raised to a certain index. or exponent.
That is
[tex] {2}^{5} = 32 \\ \\ {10}^{2} = 100[/tex]
As for the middle expression, the base and the number are equal so let us keep it for now.
Our expression now becomes,
[tex] log_{2}( {2}^{5} ) + log_{e}(e) + log_{10}( {10}^{2} )[/tex]
Recall this law of logarithm,
[tex] log_{a}( {m}^{n} ) = n log_{a}(m)[/tex]
[tex]5 log_{2}( {2}) + log_{e}(e) + 2log_{10}( {10} )[/tex]
Recall again that,
[tex] log_{a}(a) ,a \ne0 \: or \: 1[/tex]
Therefore our expression becomes,
[tex]5 (1) + (1) + 2(1) = 5 + 1 + 2 = 8[/tex]
2) We use change of base to solve this logarithmic equation.
[tex] log_{3}(x) + 4log_{9}(x) = 9[/tex]
[tex]log_{3}(x) + log_{9}( {x}^{4} ) = 9[/tex]
It will be easier and faster to change the base to 3.
Recall that,
[tex]\log_{x}(y)=\frac{log_{a}(y)}{log_{a}(x)}[/tex]
We apply this law to obtain,
[tex]log_{3}(x) + \frac{log_{3}( {x}^{4})}{log_{3}(9) } = 9[/tex]
[tex]log_{3}(x) + \frac{log_{3}( {x}^{4})}{log_{3}( {3}^{2} ) } = 9[/tex]
[tex]log_{3}(x) + \frac{log_{3}( {x}^{4})}{2log_{3}( {3} ) } = 9[/tex]
[tex]log_{3}(x) + \frac{log_{3}( {x}^{4})}{2(1) } = 9[/tex]
[tex]log_{3}(x) + \frac{log_{3}( {x}^{4})}{2} = 9[/tex]
multiplying through by 2 gives,
[tex]2log_{3}(x) + log_{3}( {x}^{4} ) = 18[/tex]
[tex]log_{3}( {x}^{2} ) + log_{3}( {x}^{4} ) = 18[/tex]
[tex] log_{3}( {x}^{4} \times {x}^{2} ) = 18[/tex]
We apply the multiplication law of exponents to obtain,
[tex] log_{3}( {x}^{4 + 2} ) = 18[/tex]
[tex] log_{3}( {x}^{6} ) = 18[/tex]
We take antilogarithms to get,
[tex] {x}^{6} = {3}^{18} [/tex]
[tex] {x}^{6} = ( {3}^{3} ) ^{6} [/tex]
[tex]x = {3}^{3} [/tex]
[tex]x = 27[/tex]
Hence x is 27.
