Solve log base 2 of x, plus log base 3 of x, equals 5.

Step-by-step solution

1. 1

   Put both logarithms in the same base. Logs in different bases cannot be added directly, but change of base fixes that:

   $\log_2 x=\frac{\ln x}{\ln 2},\qquad \log_3 x=\frac{\ln x}{\ln 3}$

   (Any common base would work; natural logs keep the notation short.)

2. 2

   Substitute $t=\ln x$ to reveal a linear equation.

   $\frac{t}{\ln 2}+\frac{t}{\ln 3}=5$

   The unknown appears only through $t$, and only to the first power — the equation is linear, not logarithmic, once written this way.

3. 3

   Factor $t$ out and combine the fractions.

   $t\left(\frac{1}{\ln 2}+\frac{1}{\ln 3}\right)=5\quad\Rightarrow\quad t\cdot\frac{\ln 3+\ln 2}{\ln 2\,\ln 3}=5\quad\Rightarrow\quad t=\frac{5\ln 2\,\ln 3}{\ln 2+\ln 3}$

   Note $\ln 2+\ln 3=\ln 6$, so equivalently $t=\dfrac{5\ln 2\,\ln 3}{\ln 6}$.

4. 4

   Evaluate $t$. With $\ln 2=0.6931472$ and $\ln 3=1.0986123$:

   $\ln 2\,\ln 3=0.7615000,\qquad \ln 6=1.7917595,\qquad t=\frac{3.8075001}{1.7917595}=2.1250062$

5. 5

   Exponentiate to recover $x$.

   $x=e^{t}=e^{2.1250062}=8.3729497$

   Since $x>1$ and both logs are positive, this is consistent — and it must exceed $2^{5/2}\approx 5.66$, because $\log_2x$ alone would need that much if $\log_3 x$ were zero.

6. 6

   Verify, and reject the value 6.06. $\log_2 8.3729497=3.0657$ and $\log_3 8.3729497=1.9343$, summing to $5.0000$ ✓. By contrast $x=6.06$ gives $2.5993+1.6400=4.2393$, nearly $0.76$ short of $5$, so it is not a solution.