Log Questions: expand log(x(y(z)^1/2)^1/2)^1/2

[jc0517]

Hey guys,

I would really appreciate it if anyone could guide me through these two questions.

1) Use log laws to expand log(x(y(z)^1/2)^1/2)^1/2

2) Without using a calculator, find an integer of x such that:
x < log9 10 < x + 1


Thanks guys!

 

[ksdhart2]

For the first problem, a review of the logarithm rules would be a good place to start, particularly noting the "Logarithm Power Rule" which says: $\displaystyle ln(x^y) = y \cdot ln(x)$. Try applying that to the outermost square root of the given logarithm. What do you get? Where does that suggest you might go next?

For the second problem, assuming that "logs 10" means the natural log (i.e. logarithm base e) of 10, you can find that value that easily enough using your calculator. What value of x does that suggest?

If you get stuck again, that's alright, but when you reply back, please re-read and comply with the rules as laid out in the Read Before Postinghttps://www.freemathhelp.com/forum/threads/41536-Read-Before-Posting!! thread that's stickied at the topic of every subforum, and share with us all the work you've done on these problems, even the parts you know for sure are wrong. Thank you.

 

[JeffM]

Hey guys,

I would really appreciate it if anyone could guide me through these two questions.

1) Use log laws to expand log(x(y(z)^1/2)^1/2)^1/2

2) Without using a calculator, find an integer of x such that:
x < log9 10 < x + 1


Thanks guys!

I agree with post # 2 preceding this one, particularly the point about reading "READ BEFORE POSTING."

With to respect to problem 1, do you mean:

$\displaystyle log \left (\sqrt{x\sqrt{y\sqrt{z}}}\right ) \text { or } \sqrt{log \left ( x\sqrt{y\sqrt{z}} \right )}.$

With respect to problem 2,

$\displaystyle log_9(1) \text { = what, and } log_9(9) \text { = what, and } log_9(81) \text { = what?}$

 

[jc0517]

My apologies for not showing my work first.
[FONT=MathJax_Math-italic]The question is this : l[/FONT][FONT=MathJax_Math-italic]o[/FONT][FONT=MathJax_Math-italic]g[/FONT][FONT=MathJax_Size2]([/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Math-italic]y[/FONT][FONT=MathJax_Math-italic]z[/FONT][FONT=MathJax_Main]√[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Size1]√[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Size2]√[/FONT][FONT=MathJax_Size2])
My working out so far is:

log x^1/2 (y (z)^1/2 ) ^1/4
log x^1/2 + log y^1/4 (z^1/8)
log x^1/2 + log y^1/4 + log z^1/8

=(log x)/2 + (log y)/4 + (log z)/8

And as for the second question, there was a typo; the question should read:
x < log3 90 < x+1 , without using a calculator

My working out for this so far is:
since log390 = log33 + log33 + log310
x < 2 + log310 < x+1
Writing the inequalities separately:
x<2 + log310 and x>1+log310
so,
1+log310 < x < 2 + log310

But this doesn't really answer the question I guess.

Any help would be greatly appreciated guys [/FONT]

 

[stapel]

...[FONT=MathJax_Size2]for the second question, there was a typo; the question should read:
x < log3 90 < x+1 , without using a calculator[/FONT]

Is the middle expression meant to say "the log, base 3, of 90"? I will assume so.

[FONT=MathJax_Size2]My working out for this so far is:
since log390 = log33 + log33 + log310
x < 2 + log310 < x+1[/FONT]

This looks good. Now, think about what logs are. They're powers. The expression, log3(10), means "the power which, when put on 3, creates an expression equal to 10". You know that 32 = 9 and 33 = 27, so clearly the power log3(10) must be between 2 and 3, since 3, raised to that power, gives you a value between 32 = 9 and 33 = 27. Where does this lead? [FONT=MathJax_Size2][/FONT]