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.

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- Nov 1st 2012, 04:44 PMGaryOgAdd/Sub Logs of different bases.
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. - Nov 1st 2012, 04:59 PMTheEmptySetRe: Add/Sub Logs of different bases.
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 - Nov 1st 2012, 05:31 PMSorobanRe: Add/Sub Logs of different bases.
Hello, GaryOg!

Quote:

$\displaystyle \text{Solve for }x\!:\;\;\log_2x + \log_3x \:=\: 1$

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: .$\displaystyle x \;=\;e^{0.425} \;=\;1.52859... \;\approx\;1.53$