Solving x in log form

[grapz]

Log 81 x + log 9 x + log 3 x = 7


the bases are 81, 9 , and 3 respectively.

Solve for X

i tried dividing each side by their logs but i can't solve for x

 

[skeeter]

\(\displaystyle \L \log_{81}x = \frac{\log_3 x}{\log_3 81} = \frac{\log_3 x}{4}\)

\(\displaystyle \L \log_{9}x = \frac{\log_3 x}{\log_3 9} = \frac{\log_3 x}{2}\)

can you finish now?

 

[grapz]

well

after that i just did this

1/4 log 3 x + 1/2 log 3 x + log 3 x = 7

then i do

log 3 ( x ^ (1/4) + x ( 1/2) + x ) = 7

which is 3 ^ 7 = x^ 1/4 + x ^ 1/2 + x

but then how do i solve for x?

 

[Mrspi]

well

after that i just did this

1/4 log 3 x + 1/2 log 3 x + log 3 x = 7

then i do

log 3 ( x ^ (1/4) + x ( 1/2) + x ) = 7

which is 3 ^ 7 = x^ 1/4 + x ^ 1/2 + x

but then how do i solve for x?

You have

(1/4) log<SUB>3</SUB> x + (1/2) log<SUB>3</SUB> x + log<SUB>3</SUB> x = 7

What if you had (1/4)a + (1/2)a + a = 7? Wouldn't you just combine like terms? Though your equation looks a bit more complicated, it is really just a situation where you combine like terms:

[(1/4) + (1/2) + 1] log<SUB>3</SUB> x = 7

(7/4) log<SUB>3</SUB> x = 7

Now....can you finish?

 

[soroban]

Hello, grapz!


You had: $\displaystyle \:\frac{1}{4}\cdot\log_3(x) \,+\, \frac{1}{2}\cdot\log_3(x) \,+\, \log_3(x)\;=\;7$

Multiply by 4: $\displaystyle \:\log_3(x)\,+\,2\cdot\log_3(x)\,+\,4\cdot\log_3(x)\;=\;28$

Then we have: $\displaystyle \:7\cdot\log_3(x) \:=\:28$

Divide by 7: $\displaystyle \:\log_3(x) \:=\:4$

Therefore: $\displaystyle \:x \;=\;3^4\;\;\Rightarrow\;\;\fbox{x\:=\:81}$

 

[soroban]

Hello, grapz!

If you don't know the Base-Change Formula, it can still be solved,
. . but it takes some gymnastics . . .

$\displaystyle \log_{81}(x)\,+\,\log_9(x)\,+\,\log_3(x)\;=\;7$

Let \(\displaystyle \log_{81}(x)\:=\\) . . . Then: $\displaystyle \:81^P \:=\:x$

. . We have: \(\displaystyle \3^4)^P\:=\:x\;\;\Rightarrow\;\;3^{4P} \:=\:x\)

. . Rewrite in log form: $\displaystyle \:4P \:=\:\log_3(x)\;\;\rightarrow\;\;P \:=\:\frac{1}{4}\cdot\log_3(x)$

. . Hence: $\displaystyle \:\log_{81}(x)\;=\;\frac{1}{4}\cdot\log_3(x)\;$ [1]


Let $\displaystyle \log_{9}(x)\:=\:Q$ . . . Then: $\displaystyle \:9^Q \:=\:x$

. . We have: \(\displaystyle \3^2)^Q\:=\:x\;\;\Rightarrow\;\;3^{2Q} \:=\:x\)

. . Rewrite in log form: $\displaystyle \:2Q \:=\:\log_3(x)\;\;\rightarrow\;\;Q \:=\:\frac{1}{2}\cdot\log_3(x)$

. . Hence: $\displaystyle \:\log_{9}(x)\;=\;\frac{1}{2}\cdot\log_3(x)\;$ [2]


Substitute [1] and [2] into the original equation:

. . . $\displaystyle \frac{1}{4}\cdot\log_3(x)\,+\,\frac{1}{2}\cdot\log_3(x)\,+\,\log_x(x)\;=\;7\;\;$ . . . see?