Question

1)log2 32+lne-lg100 2) log3 x + 4 log9 x= 9

Answer 1

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

Answer 2

To simplify we need recall and apply the properties and laws of logarithms.

1)

$log_{2}(32) + ln(e) + log_{10}(100)$

$log_{2}(32) + log_{e}(e) + log_{10}(100)$

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
${2}^{5} = 32 \\ \\ {10}^{2} = 100$

As for the middle expression, the base and the number are equal so let us keep it for now.

Our expression now becomes,

$log_{2}( {2}^{5} ) + log_{e}(e) + log_{10}( {10}^{2} )$

Recall this law of logarithm,

$log_{a}( {m}^{n} ) = n log_{a}(m)$

$5 log_{2}( {2}) + log_{e}(e) + 2log_{10}( {10} )$

Recall again that,

$log_{a}(a) ,a \ne0 \: or \: 1$

Therefore our expression becomes,
$5 (1) + (1) + 2(1) = 5 + 1 + 2 = 8$

2) We use change of base to solve this logarithmic equation.

$log_{3}(x) + 4log_{9}(x) = 9$

$log_{3}(x) + log_{9}( {x}^{4} ) = 9$

It will be easier and faster to change the base to 3.
Recall that,

$\log_{x}(y)=\frac{log_{a}(y)}{log_{a}(x)}$

We apply this law to obtain,

$log_{3}(x) + \frac{log_{3}( {x}^{4})}{log_{3}(9) } = 9$

$log_{3}(x) + \frac{log_{3}( {x}^{4})}{log_{3}( {3}^{2} ) } = 9$

$log_{3}(x) + \frac{log_{3}( {x}^{4})}{2log_{3}( {3} ) } = 9$

$log_{3}(x) + \frac{log_{3}( {x}^{4})}{2(1) } = 9$

$log_{3}(x) + \frac{log_{3}( {x}^{4})}{2} = 9$

multiplying through by 2 gives,

$2log_{3}(x) + log_{3}( {x}^{4} ) = 18$

$log_{3}( {x}^{2} ) + log_{3}( {x}^{4} ) = 18$

$log_{3}( {x}^{4} \times {x}^{2} ) = 18$

We apply the multiplication law of exponents to obtain,

$log_{3}( {x}^{4 + 2} ) = 18$

$log_{3}( {x}^{6} ) = 18$
We take antilogarithms to get,

${x}^{6} = {3}^{18}$

${x}^{6} = ( {3}^{3} ) ^{6}$

$x = {3}^{3}$
$x = 27$

Hence x is 27.