Solving x in log form
- Thread starter grapz
- Start date
grapz said:
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?
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: \(\displaystyle \:x \;=\;3^4\;\;\Rightarrow\;\;\fbox{x\:=\:81}\)
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: \(\displaystyle \:x \;=\;3^4\;\;\Rightarrow\;\;\fbox{x\:=\:81}\)
Hello, grapz!
If you don't know the Base-Change Formula, it can still be solved,
. . but it takes some gymnastics . . .
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?
If you don't know the Base-Change Formula, it can still be solved,
. . but it takes some gymnastics . . .
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?