Help with This equation. Logs
- Thread starter rawr
- Start date
I think we can solve this algebraically.
\(\displaystyle 5^{5-3x}=2^{x+2}\)
\(\displaystyle (5-3x)ln(5)=(x+2)ln(2)\)
\(\displaystyle \frac{5-3x}{x+2}=\frac{ln(2)}{ln(5)}\)
\(\displaystyle \frac{11}{x+2}-3=\frac{ln(2)}{ln(5)}\)
Add 3, then reciprocal of each side:
\(\displaystyle \frac{x+2}{11}=\frac{ln(5)}{ln(250)}\)
Now, finish up?.
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Another way:
\(\displaystyle \frac{5^{5}}{5^{3x}}=2^{2}\cdot 2^{x}\)
\(\displaystyle \frac{3125}{5^{3x}}=4\cdot 2^{x}\)
\(\displaystyle \frac{3125}{4}=(5^{3}\cdot 2)^{x}\)
Now, finish?.
\(\displaystyle 5^{5-3x}=2^{x+2}\)
\(\displaystyle (5-3x)ln(5)=(x+2)ln(2)\)
\(\displaystyle \frac{5-3x}{x+2}=\frac{ln(2)}{ln(5)}\)
\(\displaystyle \frac{11}{x+2}-3=\frac{ln(2)}{ln(5)}\)
Add 3, then reciprocal of each side:
\(\displaystyle \frac{x+2}{11}=\frac{ln(5)}{ln(250)}\)
Now, finish up?.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Another way:
\(\displaystyle \frac{5^{5}}{5^{3x}}=2^{2}\cdot 2^{x}\)
\(\displaystyle \frac{3125}{5^{3x}}=4\cdot 2^{x}\)
\(\displaystyle \frac{3125}{4}=(5^{3}\cdot 2)^{x}\)
Now, finish?.
Hello, rawr!
Another approach . . .
\(\displaystyle \text{Take logs: }\;\ln\left(5^{5-3x}\right) \:=\:\ln\left(2^{x+2}\right)\)
. . . . . . .\(\displaystyle (5-3x)\!\cdot\!\ln5 \:=\
x+2)\!\cdot\!\ln2\)
. . . . .\(\displaystyle 5\!\cdot\!\ln5 - 3x\!\cdot\!\ln5 \:=\:x\!\cdot\!\ln2 + 2\!\cdot\!\ln2\)
. . . . .\(\displaystyle x\!\cdot\!\ln2 + 3x\!\cdot\!\ln5 \:=\:5\!\cdot\!\ln5 - 2\!\cdot\!\ln2\)
. . . . .\(\displaystyle x\!\cdot\!(\ln 2 + 3\ln 5) \:=\:5\!\cdot\!\ln5 - 2\!\cdot\!\ln 2\)
. . . . . . . . . . . . . .\(\displaystyle x \:=\:\frac{5\!\cdot\!\ln5 - 2\!\cdot\!\ln2}{\ln 2 + 3\!\cdot\!\ln5}\)
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
This can be simplified beyond all recognition . . .
. . \(\displaystyle x \;=\;\frac{\ln(5^5) - \ln(2^2)}{\ln(5^3) + \ln2} \;=\; \frac{\ln3125 - \ln4}{\ln125 + \ln 2}\)
. . . . \(\displaystyle = \;\frac{\ln\left(\frac{3125}{4}\right)}{\ln(125\cdot2)} \;=\;\frac{\ln781.25}{\ln250}\)
. . . . \(\displaystyle = \;1.206364638\hdots\)
Another approach . . .
\(\displaystyle \text{Take logs: }\;\ln\left(5^{5-3x}\right) \:=\:\ln\left(2^{x+2}\right)\)
. . . . . . .\(\displaystyle (5-3x)\!\cdot\!\ln5 \:=\
x+2)\!\cdot\!\ln2\). . . . .\(\displaystyle 5\!\cdot\!\ln5 - 3x\!\cdot\!\ln5 \:=\:x\!\cdot\!\ln2 + 2\!\cdot\!\ln2\)
. . . . .\(\displaystyle x\!\cdot\!\ln2 + 3x\!\cdot\!\ln5 \:=\:5\!\cdot\!\ln5 - 2\!\cdot\!\ln2\)
. . . . .\(\displaystyle x\!\cdot\!(\ln 2 + 3\ln 5) \:=\:5\!\cdot\!\ln5 - 2\!\cdot\!\ln 2\)
. . . . . . . . . . . . . .\(\displaystyle x \:=\:\frac{5\!\cdot\!\ln5 - 2\!\cdot\!\ln2}{\ln 2 + 3\!\cdot\!\ln5}\)
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
This can be simplified beyond all recognition . . .
. . \(\displaystyle x \;=\;\frac{\ln(5^5) - \ln(2^2)}{\ln(5^3) + \ln2} \;=\; \frac{\ln3125 - \ln4}{\ln125 + \ln 2}\)
. . . . \(\displaystyle = \;\frac{\ln\left(\frac{3125}{4}\right)}{\ln(125\cdot2)} \;=\;\frac{\ln781.25}{\ln250}\)
. . . . \(\displaystyle = \;1.206364638\hdots\)
D
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Denis said:
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