logarithm eqn: need help w/ log_x(64) + log_8x(32) = 4
- Thread starter oded244
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skeeter
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Re: logarithm ...
seems more likely that it's
\(\displaystyle \log_x(64) + \log_{8x}(32) = 4\)
change to base 2 ...
\(\displaystyle \frac{\log_2(64)}{\log_2(x)} + \frac{\log_2(32)}{\log_2(8x)} = 4\)
\(\displaystyle \frac{6}{\log_2(x)} + \frac{5}{\log_2(8) + \log_2(x)} = 4\)
\(\displaystyle \frac{6}{\log_2(x)} + \frac{5}{3 + \log_2(x)} = 4\)
let
\(\displaystyle t = \log_2(x)\)
\(\displaystyle \frac{6}{t} + \frac{5}{3 + t} = 4\)
\(\displaystyle \frac{6(3 + t)}{t(3 + t)} + \frac{5t}{t(3 + t)} = 4\)
\(\displaystyle \frac{18+11t}{t(3+t)} = 4\)
\(\displaystyle 18+11t = 12t + 4t^2\)
\(\displaystyle 0 = 4t^2 + t - 18\)
\(\displaystyle 0 = (t-2)(4t+9)\)
\(\displaystyle t = 2\)
\(\displaystyle x = 4\)
\(\displaystyle t = -\frac{9}{4}\)
\(\displaystyle x = 2^{-\frac{9}{4}}\) ... throw this solution out, base x must be > 0.
seems more likely that it's
\(\displaystyle \log_x(64) + \log_{8x}(32) = 4\)
change to base 2 ...
\(\displaystyle \frac{\log_2(64)}{\log_2(x)} + \frac{\log_2(32)}{\log_2(8x)} = 4\)
\(\displaystyle \frac{6}{\log_2(x)} + \frac{5}{\log_2(8) + \log_2(x)} = 4\)
\(\displaystyle \frac{6}{\log_2(x)} + \frac{5}{3 + \log_2(x)} = 4\)
let
\(\displaystyle t = \log_2(x)\)
\(\displaystyle \frac{6}{t} + \frac{5}{3 + t} = 4\)
\(\displaystyle \frac{6(3 + t)}{t(3 + t)} + \frac{5t}{t(3 + t)} = 4\)
\(\displaystyle \frac{18+11t}{t(3+t)} = 4\)
\(\displaystyle 18+11t = 12t + 4t^2\)
\(\displaystyle 0 = 4t^2 + t - 18\)
\(\displaystyle 0 = (t-2)(4t+9)\)
\(\displaystyle t = 2\)
\(\displaystyle x = 4\)
\(\displaystyle t = -\frac{9}{4}\)
\(\displaystyle x = 2^{-\frac{9}{4}}\) ... throw this solution out, base x must be > 0.
skeeter
Elite Member
- Joined
- Dec 15, 2005
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Re: logarithm ...
\(\displaystyle \log_2(x) + \log_{2x}(x^3) = 4\)
\(\displaystyle \log_2(x) + \frac{\log_2(x^3)}{\log_2(2x)} = 4\)
\(\displaystyle \log_2(x) + \frac{3\log_2(x)}{\log_2(2) + \log_2(x)} = 4\)
let
\(\displaystyle t = \log_2(x)\)
\(\displaystyle t + \frac{3t}{1+t} = 4\)
take it from here?
\(\displaystyle \log_2(x) + \log_{2x}(x^3) = 4\)
\(\displaystyle \log_2(x) + \frac{\log_2(x^3)}{\log_2(2x)} = 4\)
\(\displaystyle \log_2(x) + \frac{3\log_2(x)}{\log_2(2) + \log_2(x)} = 4\)
let
\(\displaystyle t = \log_2(x)\)
\(\displaystyle t + \frac{3t}{1+t} = 4\)
take it from here?
Re: logarithm ...
Hello, oded244!
\(\displaystyle \text{We have: }\;\log_x(2^6) + \log_{8x}(2^5) \;=\;4 \quad\Rightarrow\quad 6\log_x(2) + 5\log_{8x}(2) \;=\;4\)
. . \(\displaystyle \frac{6}{\log_2(x)} + \frac{5}{\log_2(8x)} \;=\;4 \quad\Rightarrow\quad 6\log_2(8x) + 5\log_2(x) \;=\;4\log_2(x)\log_2(8x)\)
. . \(\displaystyle 6\bigg[\log_2(8) + \log_2(x)\bigg] + 5\log_2(x) \;=\;4\log_2(x)\bigg[\log_2(8) + \log_2(x)\bigg]\)
. . \(\displaystyle 6\cdot 3 + 6\log_2(x) + 5\log_2(x) \;=\;4\log_2(x)\bigg[3 + \log_2(x)\bigg]\)
. . \(\displaystyle 18 + 11\log_2(x) \;=\;12\log_2(x) + 4\left[\log_2(x)\right]^2\)
\(\displaystyle \text{We have the quadratic: }\;4\left[\log_2(x)\right]^2 + \log_2(x) - 18 \;=\;0\)
. . \(\displaystyle \text{which factors: }\;\left[4\log_2(x) + 9\right]\left[\log_2(x) - 2\right] \;=\;0\)
Therefore . . .
\(\displaystyle 4\log_2(x) + 9 \:=\:0 \quad\Rightarrow\quad \log_2(x) \:=\:-\frac{9}{4} \quad\Rightarrow\quad\boxed{ x \;=\;2^{-\frac{9}{4}}}\)
\(\displaystyle \log_2(x) - 2 \:=\:0 \quad\Rightarrow\quad \log_2(x) \:=\:2 \quad\Rightarrow\quad x \:=\:2^2 \quad\Rightarrow\quad\boxed{x\:=\:4}\)
Hello, oded244!
\(\displaystyle \text{We have: }\;\log_x(2^6) + \log_{8x}(2^5) \;=\;4 \quad\Rightarrow\quad 6\log_x(2) + 5\log_{8x}(2) \;=\;4\)
. . \(\displaystyle \frac{6}{\log_2(x)} + \frac{5}{\log_2(8x)} \;=\;4 \quad\Rightarrow\quad 6\log_2(8x) + 5\log_2(x) \;=\;4\log_2(x)\log_2(8x)\)
. . \(\displaystyle 6\bigg[\log_2(8) + \log_2(x)\bigg] + 5\log_2(x) \;=\;4\log_2(x)\bigg[\log_2(8) + \log_2(x)\bigg]\)
. . \(\displaystyle 6\cdot 3 + 6\log_2(x) + 5\log_2(x) \;=\;4\log_2(x)\bigg[3 + \log_2(x)\bigg]\)
. . \(\displaystyle 18 + 11\log_2(x) \;=\;12\log_2(x) + 4\left[\log_2(x)\right]^2\)
\(\displaystyle \text{We have the quadratic: }\;4\left[\log_2(x)\right]^2 + \log_2(x) - 18 \;=\;0\)
. . \(\displaystyle \text{which factors: }\;\left[4\log_2(x) + 9\right]\left[\log_2(x) - 2\right] \;=\;0\)
Therefore . . .
\(\displaystyle 4\log_2(x) + 9 \:=\:0 \quad\Rightarrow\quad \log_2(x) \:=\:-\frac{9}{4} \quad\Rightarrow\quad\boxed{ x \;=\;2^{-\frac{9}{4}}}\)
\(\displaystyle \log_2(x) - 2 \:=\:0 \quad\Rightarrow\quad \log_2(x) \:=\:2 \quad\Rightarrow\quad x \:=\:2^2 \quad\Rightarrow\quad\boxed{x\:=\:4}\)
Re: logarithm ...
Hello again, oded244!
\(\displaystyle \text{We have: }\;\log_2(x) + 3\log_{2x}(x) \;=\;4\)
. . \(\displaystyle \frac{1}{\log_x(2)} + \frac{3}{\log_x(2x)} \;=\;4\)
. . \(\displaystyle \log_x(2x) + 3\log_x(2) \;=\;4\log_x(2)\log_x(2x)\)
. . \(\displaystyle \log_x(2) + \log_x(x) + 3\log_x(2) \;=\;4\log_x(2)\bigg[\log_x(2) + \log_x(x)\bigg]\)
. . \(\displaystyle 4\log_x(2) + 1 \;=\;4\log_x(2)\bigg[\log_x(2) + 1\bigg]\)
. . \(\displaystyle 4\log_x(2) + 1 \;=\;4\bigg[\log_x(2)\bigg]^2 + 4\log_x(2)\)
\(\displaystyle \text{We have: }\;4\bigg[\log_x(2)\bigg]^2 \;=\;1 \quad\Rightarrow\quad\bigg[\log_x(2)\bigg]^2 \;=\;\frac{1}{4} \quad\Rightarrow\quad \log_x(2) \;=\;\pm\frac{1}{2}\)
. . \(\displaystyle \log_x(2) \:=\:\frac{1}{2} \quad\Rightarrow\quad x^{\frac{1}{2}} \:=\:2 \quad\Rightarrow\quad\boxed{ x \:=\:4}\)
. . \(\displaystyle \log_x(2) \:=\:-\frac{1}{2} \quad\Rightarrow\quad x^{-\frac{1}{2}} \:=\:2 \quad\Rightarrow\quad \boxed{x \:=\:\frac{1}{4}}\)
Hello again, oded244!
\(\displaystyle \text{We have: }\;\log_2(x) + 3\log_{2x}(x) \;=\;4\)
. . \(\displaystyle \frac{1}{\log_x(2)} + \frac{3}{\log_x(2x)} \;=\;4\)
. . \(\displaystyle \log_x(2x) + 3\log_x(2) \;=\;4\log_x(2)\log_x(2x)\)
. . \(\displaystyle \log_x(2) + \log_x(x) + 3\log_x(2) \;=\;4\log_x(2)\bigg[\log_x(2) + \log_x(x)\bigg]\)
. . \(\displaystyle 4\log_x(2) + 1 \;=\;4\log_x(2)\bigg[\log_x(2) + 1\bigg]\)
. . \(\displaystyle 4\log_x(2) + 1 \;=\;4\bigg[\log_x(2)\bigg]^2 + 4\log_x(2)\)
\(\displaystyle \text{We have: }\;4\bigg[\log_x(2)\bigg]^2 \;=\;1 \quad\Rightarrow\quad\bigg[\log_x(2)\bigg]^2 \;=\;\frac{1}{4} \quad\Rightarrow\quad \log_x(2) \;=\;\pm\frac{1}{2}\)
. . \(\displaystyle \log_x(2) \:=\:\frac{1}{2} \quad\Rightarrow\quad x^{\frac{1}{2}} \:=\:2 \quad\Rightarrow\quad\boxed{ x \:=\:4}\)
. . \(\displaystyle \log_x(2) \:=\:-\frac{1}{2} \quad\Rightarrow\quad x^{-\frac{1}{2}} \:=\:2 \quad\Rightarrow\quad \boxed{x \:=\:\frac{1}{4}}\)