# Add/Sub Logs of different bases.

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• November 1st 2012, 05:44 PM
GaryOg
Add/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.
• November 1st 2012, 05:59 PM
TheEmptySet
Re: Add/Sub Logs of different bases.
Quote:

Originally Posted by GaryOg
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

$2^{\frac{\log(3)}{\log(6)}} \text{ or }3^{\frac{\log(2)}{\log(6)}}$

I would try by converting everything to a base 10 logarithm

$\frac{\log(x)}{\log(2)}+\frac{\log(x)}{\log(3)}=1 \iff \log x^{\log 3}+x^{\log 2}=\log 2^{\log 3}$

Now try to use the log properties to finish
• November 1st 2012, 06:31 PM
Soroban
Re: Add/Sub Logs of different bases.
Hello, GaryOg!

Quote:

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

I used the Base-Change Formula: . $\log_ba \:=\:\frac{\ln a}{\ln b}$

The equation becomes: . $\frac{\ln x}{\ln2} + \frac{\ln x}{\ln3} \:=\:1$

Multiply by $\ln2\!\cdot\!\ln3\!:$

. . $\ln3\!\cdot\!\ln x + \ln2\!\cdot\!\ln x \:=\:\ln2\ln3$

. . . . . . $(\ln2 + \ln3)\ln x \:=\:\ln2\!\cdot\!\ln3$

. . . . . . . . . . . . . . $\ln x \:=\:\frac{\ln2\!\cdot\!\ln3}{\ln2 + \ln3} \;=\;\frac{\ln2\!\cdot\!\ln3}{\ln6} \;\approx\;0.425$

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