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

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

Find x

(a) 8 (b) 16 (c) 4 (d) 32

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

- stapel_eliz
**Posts:**1738**Joined:**Mon Dec 08, 2008 4:22 pm-
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stapel_eliz wrote:partap95 wrote: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
**Posts:**1738**Joined:**Mon Dec 08, 2008 4:22 pm-
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partap95 wrote: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?

. . . . .

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!

stapel_eliz wrote:partap95 wrote: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?

. . . . .

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
**Posts:**202**Joined:**Mon Dec 08, 2008 5:18 pm-
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stapel_eliz wrote:So the equation is as follows?

. . . . .

partap95 wrote: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

little_dragon wrote:stapel_eliz wrote:So the equation is as follows?

. . . . .partap95 wrote: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?

partap95 wrote: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?