If log[sub:3fcs7lcy]9[/sub:3fcs7lcy]5=x and log[sub:3fcs7lcy]27[/sub:3fcs7lcy]6=y, find an expression in terms of x and y for log[sub:3fcs7lcy]3[/sub:3fcs7lcy]1.2 I know it has something to do with base 3 ,and I have tried to write it in exp. form: 3[sup:3fcs7lcy]2x[/sup:3fcs7lcy]=5, 3[sup:3fcs7lcy]3y[/sup:3fcs7lcy] = 6 , 3[sup:3fcs7lcy]x[/sup:3fcs7lcy] =1.2 I don't know which direction to go next. Does the question mean isolate and solve for x? Does it mean substitute 3[sup:3fcs7lcy]x[/sup:3fcs7lcy] = 1.2 into the other 2 questions somehow? Thanks for any help you can give.
logarithm question
- Thread starter gkagawa
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Welcome back pka - I was missing that quote from Auden - now I can go to sleep.
Hello, gkagawa!
I'll solve it baby-steps for you . . .
\(\displaystyle \text{We have: }\:\log_95 \:=\:x \quad\Rightarrow\quad 9^x \:=\:5 \quad\Rightarrow\quad (3^2)^x \:=\:5 \quad\Rightarrow\quad 3^{2x} \:=\:5\)
\(\displaystyle \text{Take logs, base 3: }\:\log_3(3^{2x}) \:=\:\log_35 \quad\Rightarrow\quad 2x\cdot\underbrace{\log_33}_{\text{This is 1}} \:=\:\log_35 \quad\Rightarrow\quad\boxed{ 2x \:=\:\log_35}\;\;[1]\)
\(\displaystyle \text{We have: }\:\log_{27}6 \,=\,y \quad\Rightarrow\quad 27^y \:=\:6 \quad\Rightarrow\quad (3^3)^y \:=\:6 \quad\Rightarrow\quad 3^{3y} \:=\:6\)
\(\displaystyle \text{Take logs, base 3: }\:\log_3(3^{3y}) \:=\:\log_56 \quad\Rightarrow\quad 3y\cdot\underbrace{\log_33}_{\text{This is 1}} \:=\:\log_56 \quad\Rightarro\quad \boxed{3y \:=\:\log_36}\;\;[2]\)
\(\displaystyle \text{Note that: }\:\log_3(1.2) \;=\;\log_3\left(\frac{6}{5}\right) \;=\;\log_36 - \log_35\)
\(\displaystyle \text{Substitute [1] and [2]: }\;\log(1.2) \;=\;3y - 2x\)
I'll solve it baby-steps for you . . .
\(\displaystyle \text{We have: }\:\log_95 \:=\:x \quad\Rightarrow\quad 9^x \:=\:5 \quad\Rightarrow\quad (3^2)^x \:=\:5 \quad\Rightarrow\quad 3^{2x} \:=\:5\)
\(\displaystyle \text{Take logs, base 3: }\:\log_3(3^{2x}) \:=\:\log_35 \quad\Rightarrow\quad 2x\cdot\underbrace{\log_33}_{\text{This is 1}} \:=\:\log_35 \quad\Rightarrow\quad\boxed{ 2x \:=\:\log_35}\;\;[1]\)
\(\displaystyle \text{We have: }\:\log_{27}6 \,=\,y \quad\Rightarrow\quad 27^y \:=\:6 \quad\Rightarrow\quad (3^3)^y \:=\:6 \quad\Rightarrow\quad 3^{3y} \:=\:6\)
\(\displaystyle \text{Take logs, base 3: }\:\log_3(3^{3y}) \:=\:\log_56 \quad\Rightarrow\quad 3y\cdot\underbrace{\log_33}_{\text{This is 1}} \:=\:\log_56 \quad\Rightarro\quad \boxed{3y \:=\:\log_36}\;\;[2]\)
\(\displaystyle \text{Note that: }\:\log_3(1.2) \;=\;\log_3\left(\frac{6}{5}\right) \;=\;\log_36 - \log_35\)
\(\displaystyle \text{Substitute [1] and [2]: }\;\log(1.2) \;=\;3y - 2x\)