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.
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
Hello, GaryOg!
$\displaystyle \text{Solve for }x\!:\;\;\log_2x + \log_3x \:=\: 1$
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: .$\displaystyle x \;=\;e^{0.425} \;=\;1.52859... \;\approx\;1.53$