## help w/ logs: if [log_x(1/2) 2]^2 = log_x 2 Find x

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partap95
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### help w/ logs: if [log_x(1/2) 2]^2 = log_x 2 Find x

if [log_x(1/2) 2]^2 = log_x 2
Find x
(a) 8 (b) 16 (c) 4 (d) 32

the whole log is being squared.. and x(1/2) = underoot x

stapel_eliz
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if [log_x(1/2) 2]^2 = log_x 2
Find x
(a) 8 (b) 16 (c) 4 (d) 32

the whole log is being squared.. and x(1/2) = underoot x
I've heard of a log having a "base", but I've never heard of an "underoot". Please reply with your book's definition of "underoot". Thank you!

partap95
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### Re:

if [log_x(1/2) 2]^2 = log_x 2
Find x
(a) 8 (b) 16 (c) 4 (d) 32

the whole log is being squared.. and x(1/2) = underoot x
I've heard of a log having a "base", but I've never heard of an "underoot". Please reply with your book's definition of "underoot". Thank you!
The base of the first log is x(1/2). Ignore the underoot.(it just refers to the power 1/2). Please help me with this. Thanks

stapel_eliz
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The base of the first log is x(1/2). Ignore the underoot.(it just refers to the power 1/2).
Do you mean that the base of the logarithm on the left-hand side is "the square root of x"? Is "x" he base of the log on the right-hand side? So the equation is as follows?

. . . . .$\left(\log_{\sqrt{x}}(2)\right)^2\, =\, \log_x (2)$

When you reply, please show what you've tried so far, even if it was just plugging the various answer-options into the equation to see if they work. Thank you!

partap95
Posts: 5
Joined: Mon Oct 07, 2013 10:57 am
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### Re:

The base of the first log is x(1/2). Ignore the underoot.(it just refers to the power 1/2).
Do you mean that the base of the logarithm on the left-hand side is "the square root of x"? Is "x" he base of the log on the right-hand side? So the equation is as follows?

. . . . .$\left(\log_{\sqrt{x}}(2)\right)^2\, =\, \log_x (2)$

When you reply, please show what you've tried so far, even if it was just plugging the various answer-options into the equation to see if they work. Thank you!
Yes that's the equation. I just plugged in the answers with 16 being the correct answer. Is there not any other way to solve this?

little_dragon
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So the equation is as follows?

. . . . .$\left(\log_{\sqrt{x}}(2)\right)^2\, =\, \log_x (2)$
Yes that's the equation. I just plugged in the answers with 16 being the correct answer. Is there not any other way to solve this?
i think U could use base change
log_sqrt[x](2)=log_x(2) / log_x(sqrt[x])=log_x(2) / (1/2)log_x(x)
& log_x(x)=1 so
log_x(2)/(1/2)=2log_x(2)
sqrd is 4(log_x(2))
then eqn is 4(log_x(2))^2=log_x(2)
4(log_x(2))^2-log_x(2)=0
log_x(2) [4log_x(2)-1]=0
log_x(2)=0 => x=1 but base cant be 1
4log_x(2)-1=0 => log_x(2)=1/4
x^(1/4)=2 => x=2^4=16

plz rite back if that doesnt make sense
thnx

partap95
Posts: 5
Joined: Mon Oct 07, 2013 10:57 am
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### Re:

So the equation is as follows?

. . . . .$\left(\log_{\sqrt{x}}(2)\right)^2\, =\, \log_x (2)$
Yes that's the equation. I just plugged in the answers with 16 being the correct answer. Is there not any other way to solve this?
i think U could use base change
log_sqrt[x](2)=log_x(2) / log_x(sqrt[x])=log_x(2) / (1/2)log_x(x)
& log_x(x)=1 so
log_x(2)/(1/2)=2log_x(2)
sqrd is 4(log_x(2))
then eqn is 4(log_x(2))^2=log_x(2)
4(log_x(2))^2-log_x(2)=0
log_x(2) [4log_x(2)-1]=0
log_x(2)=0 => x=1 but base cant be 1
4log_x(2)-1=0 => log_x(2)=1/4
x^(1/4)=2 => x=2^4=16

plz rite back if that doesnt make sense
thnx
Wonderful! Thank you so much. Also,
2log a = log b
log a^2 = log b
a^2 = b
Why cannot 2 log a = log b; 2a=b work?

buddy
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### Re: Re:

2log a = log b
log a^2 = log b
a^2 = b
Why cannot 2 log a = log b; 2a=b work?
Do a new post for a new question. Thanks.
You can't do it the way you say because the argument of the one log is "a^2", n ot "2a". It's not like "log" is a variable or a factor or anything. Its the name of the function. You wouldn't say "2*f(a)" is the same as "f(a^2)", right?

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