# Add/Sub Logs of different bases.

#### 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.

#### TheEmptySet

MHF Hall of Honor
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

1 person

#### Soroban

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

1 person