G GaryOg Jul 2012 2 0 Galway Nov 1, 2012 #1 Hello, Regarding, Solve for x, Log(base2)x + Log(base3)x = 1 I keep getting x = sq root of 6. The answer at the back of the book say x = 1.53. Is the book wrong? / what am I missing? Thank you, Gary Og.

Hello, Regarding, Solve for x, Log(base2)x + Log(base3)x = 1 I keep getting x = sq root of 6. The answer at the back of the book say x = 1.53. Is the book wrong? / what am I missing? Thank you, Gary Og.

TheEmptySet MHF Hall of Honor Feb 2008 3,764 2,029 Yuma, AZ, USA Nov 1, 2012 #2 GaryOg said: Hello, Regarding, Solve for x, Log(base2)x + Log(base3)x = 1 I keep getting x = sq root of 6. The answer at the back of the book say x = 1.53. Is the book wrong? / what am I missing? Thank you, Gary Og. Click to expand... Since I don't know what you did, I can't say where you went wrong. I get the exact answer as \(\displaystyle 2^{\frac{\log(3)}{\log(6)}} \text{ or }3^{\frac{\log(2)}{\log(6)}}\) I would try by converting everything to a base 10 logarithm \(\displaystyle \frac{\log(x)}{\log(2)}+\frac{\log(x)}{\log(3)}=1 \iff \log x^{\log 3}+x^{\log 2}=\log 2^{\log 3}\) Now try to use the log properties to finish Reactions: 1 person

GaryOg said: Hello, Regarding, Solve for x, Log(base2)x + Log(base3)x = 1 I keep getting x = sq root of 6. The answer at the back of the book say x = 1.53. Is the book wrong? / what am I missing? Thank you, Gary Og. Click to expand... Since I don't know what you did, I can't say where you went wrong. I get the exact answer as \(\displaystyle 2^{\frac{\log(3)}{\log(6)}} \text{ or }3^{\frac{\log(2)}{\log(6)}}\) I would try by converting everything to a base 10 logarithm \(\displaystyle \frac{\log(x)}{\log(2)}+\frac{\log(x)}{\log(3)}=1 \iff \log x^{\log 3}+x^{\log 2}=\log 2^{\log 3}\) Now try to use the log properties to finish

S Soroban MHF Hall of Honor May 2006 12,028 6,341 Lexington, MA (USA) Nov 1, 2012 #3 Hello, GaryOg! \(\displaystyle \text{Solve for }x\!:\;\;\log_2x + \log_3x \:=\: 1\) Click to expand... I used the Base-Change Formula: .\(\displaystyle \log_ba \:=\:\frac{\ln a}{\ln b}\) The equation becomes: .\(\displaystyle \frac{\ln x}{\ln2} + \frac{\ln x}{\ln3} \:=\:1\) Multiply by \(\displaystyle \ln2\!\cdot\!\ln3\!:\) . . \(\displaystyle \ln3\!\cdot\!\ln x + \ln2\!\cdot\!\ln x \:=\:\ln2\ln3\) . . . . . .\(\displaystyle (\ln2 + \ln3)\ln x \:=\:\ln2\!\cdot\!\ln3\) . . . . . . . . . . . . . . \(\displaystyle \ln x \:=\:\frac{\ln2\!\cdot\!\ln3}{\ln2 + \ln3} \;=\;\frac{\ln2\!\cdot\!\ln3}{\ln6} \;\approx\;0.425\) Therefore: .\(\displaystyle x \;=\;e^{0.425} \;=\;1.52859... \;\approx\;1.53\) Reactions: 1 person

Hello, GaryOg! \(\displaystyle \text{Solve for }x\!:\;\;\log_2x + \log_3x \:=\: 1\) Click to expand... I used the Base-Change Formula: .\(\displaystyle \log_ba \:=\:\frac{\ln a}{\ln b}\) The equation becomes: .\(\displaystyle \frac{\ln x}{\ln2} + \frac{\ln x}{\ln3} \:=\:1\) Multiply by \(\displaystyle \ln2\!\cdot\!\ln3\!:\) . . \(\displaystyle \ln3\!\cdot\!\ln x + \ln2\!\cdot\!\ln x \:=\:\ln2\ln3\) . . . . . .\(\displaystyle (\ln2 + \ln3)\ln x \:=\:\ln2\!\cdot\!\ln3\) . . . . . . . . . . . . . . \(\displaystyle \ln x \:=\:\frac{\ln2\!\cdot\!\ln3}{\ln2 + \ln3} \;=\;\frac{\ln2\!\cdot\!\ln3}{\ln6} \;\approx\;0.425\) Therefore: .\(\displaystyle x \;=\;e^{0.425} \;=\;1.52859... \;\approx\;1.53\)