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:**1628**Joined:**Mon Dec 08, 2008 4:22 pm-
**Contact:**

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. ThanksI'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!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
**Posts:**1628**Joined:**Mon Dec 08, 2008 4:22 pm-
**Contact:**

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?The base of the first log is x(1/2). Ignore the underoot.(it just refers to the power 1/2).

. . . . .

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?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?The base of the first log is x(1/2). Ignore the underoot.(it just refers to the power 1/2).

. . . . .

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!

- little_dragon
**Posts:**226**Joined:**Mon Dec 08, 2008 5:18 pm-
**Contact:**

So the equation is as follows?

. . . . .

i think U could use base changeYes 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?

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,So the equation is as follows?

. . . . .i think U could use base changeYes 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?

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

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.2log a = log b

log a^2 = log b

a^2 = b

Why cannot 2 log a = log b; 2a=b work?

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?