Question

Help please please please please please 6th & 7th

Answer 1

First lets start with number six. The only way to solve this is if you determine what "a" and "b" are using the first log they have given to you. $log_a_ba=\frac{1}{3}$

The first variable that I solved for was "a" and $a=e^\frac{in(b)}{2}[tex]{[tex] 0$}

The same is also true for "b", but when you put both "a" and "b" together the only combination that I have found to work is $a=\frac{1}{2} , b=\frac{1}{4}$

Next you plug these numbers in for "a" and "b" on the second equation to get something that looks like this: $log_\frac{1}{2}_*_\frac{1}{4} (\frac{\sqrt{\frac{1}{2}}}{\sqrt[3]{\frac{1}{4}}} )= -\frac{1}{18}$ and the picture below shows where the answer becomes a negative fraction.

https://www.symbolab.com/solver/logarithms-calculator/%20%5Clog_%7B%5Cfrac%7B1%7D%7B2%7D%5Ccdot%5Cfrac%7B1%7D%7B4%7D%7D%5Cleft(%5Cfrac%7B%5Csqrt%7B%5Cfrac%7B1%7D%7B2%7D%7D%7D%7B%5Csqrt%5B3%5D%7B%5Cfrac%7B1%7D%7B4%7D%7D%7D%5Cright)

If you paste that link in your search bar it will give you a even more in depth understanding of how to get this answer

Next is #7, the easier of the two.There are two ways to solve for your answers. According to the graph of this equation there are four possible real solutions. $x=2,-2,\sqrt{2}, and -\sqrt{2}$ . (This does not account for any complex solutions)

Notice that the bases are conjugates which is why the answers are so "nice"

The key is in the exponents

if $x^{2} -3 =1$ then the sum on the conjugates will be 10 so

$x^{2}-3=1$
$x^{2}=4$
so $x= 2 or -2$

Now for the other two

the solution is also true if $x^{2}-3=-1$

so

$x^{2}-3 = -1\\x^{2}=2\\x=\left \{ {{\sqrt{2}} \atop {-\sqrt{2}}} \right.$

the four real solutions are $2, -2, \sqrt{2} ,-\sqrt{2}$