# Solve for x log2 (x) +log3 (x)=1.

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### 2 Answers

We have to solve for x given that log(2) x + log(3) x = 1

log(2) x + log(3) x = 1

take the log to the base 10

log x/log 2 + log x /log 3 = 1

=> log 3 * log x + log 2 * log x = log 2* log 3

=> log x ( log 2 + log 3) = log 2* log 3

=> log x = (log 2* log 3)/( log 2 + log 3)

=> log x = (log 2 * log 3)/log 6

=> x = 10^[(log 2* log 3)/log 6]

**The value of x = 10^[(log 2* log 3)/log 6]**

We'll change the bases of logarithms into the common base 10.

log2 (x) = log x/log 2

log3 (x) = log x/log 3

We'll re-write the equation:

log x/log 2 + log x/log 3 = 1

We'll calculate the LCD = (log 2)*(log 3)

We'll multiply all over by (log 2)*(log 3):

(log 2)*(log 3)*(log x)/log 2 + (log 2)*(log 3)*(log x)/log 3 = (log 2)*(log 3)

(log 3)*(log x) + (log 2)*(log x) = (log 2)*(log 3)

We'll factorize by log x:

(log x)*[(log 3) + (log 2)] = (log 2)*(log 3)

We'll apply the product property of logarithms:

log x = (log 2)*(log 3)/log(2*3)

log x = (log 2)*(log 3)/log 6

Since the base is 10, we'll take antilog and we'll get:

x = 10^[(log 2)*(log 3)/log 6]

**The solution of the given equation is x = 10^[(log 2)*(log 3)/log 6].**