Add/Sub Logs of different bases.

Jul 2012
2
0
Galway
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
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
 
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Soroban

MHF Hall of Honor
May 2006
12,028
6,341
Lexington, MA (USA)
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: .\(\displaystyle x \;=\;e^{0.425} \;=\;1.52859... \;\approx\;1.53\)
 
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