Answered

I need help on a mathematics question. I need an explanation.

Solve 2 log x = log 64.

x = 8
x = ±8
x = 32
x = 128
Why is it not B and rather A? I agree with 8, but I also think that -8 works.



Answer :

To solve this problem you must apply the proccedure shown below:

1- To solve the logarithm expression [tex] 2 log x = log 64 [/tex], you must apply the following properties:

[tex] alogx=logx^{a} [/tex]

[tex] logx-logy=log(x/y) [/tex]

[tex] a^{log(a)x}=x [/tex]

2-Keeping this on mind, you have:

[tex] log x^{2}-log 64=0\\ log(x^{2} /64)=0\\ 10^{log(x^{2} /64)} =10^{0} \\ x^{2} /64=1 [/tex]

3. Now, solve for [tex] x [/tex]:

[tex] x=\sqrt{64}\\ x=8, x=-8 [/tex]

4. If you susbtitute [tex] x=-8 [/tex] into the expression shown in the problem, you will see that it is invalid.

Therefore. the answer is: [tex] x=8 [/tex].