Question

solve the equation. log(base 2)x+log(base 9)x=1

Answer 1

\[\log_{2}X+\log_{9}X=1 \]

Answer 2

There must be a nicer way but here's what I did.

rearrange the logarithms into exponential form:
\[\large 9^a =x\]
\[\large 2^b = x\]
which leads to the equation:
\[\large a+b = 1\]
but now we need one more equation since there are now three variables and only two equations.

\[\large log_{2}x+\log_{9}x =1\]
\[\huge 2^{\log_{2}x+\log_{9}x}= 2^1\]
\[\huge 2^{\log_{2}x}*2^{\log_{9}x}=2\]
\[\huge x*2^{\log_{9}x}=2\]
\[\huge x=\frac{2}{2^{\log_9x}}\]
sub from above:
\[\large [\log_9x=a]\]
\[\huge x=\frac{2}{2^{a}}\]
\[\huge x = 2^{(1-a)}\]

Answer 3

like wise if we raise everything to 9 in the initial step we arrive at:
\[\huge x = 9(1-b)\]
since x = x we get:

\[\huge 2^{(1-a)}=9^{(1-b)}\]
this with equation:
\[\huge [a+b=1]\]
allows you to solve either a or b, after which x is readily solvable

Answer 4

Thanks

Answer 5

@ladyalice was able to clean it up a bit and make it more straight forward

Answer 6

@Ladyalice25
You can ignore the other stuff, the most direct way consists of breaking the logs using the following log identity:
\[\large \log_a b = \frac{\log_{10}b}{\log_{10}a}=\frac{\log b}{\log a}\]
you can collect the x term within a few steps and the rest is straight algebra.

\[\large \log_2x + \log_9x = 1\]
\[\large \frac{\log x}{\log 2}+\frac{\log x}{\log 9}=1\]
\[\large \log x(\frac{1}{\log2}+\frac{1}{\log9})=1\]
\[\large = \log x(\frac{\log 9 +\log 2}{\log 2 * \log 9})=1\]
\[\large \log x (\frac{\log 18}{\log 2 * \log 9})=1\]
\[\large \log x = \frac{\log 2 * \log 9}{\log 18}\]
\[\huge x = 10^{(\frac{\log 2 * \log 9}{\log 18})}\]