Question

Solve the following for x. log2x+log3=log5

Answer 1

http://www.chilimath.com/algebra/advanced/log/images/rules%20of%20exponents.gif
see rule number 1 there
can you simplify the left-hand side?

Answer 2

log 6x?

Answer 3

\[\log2x = \log 5 - \log3\]
\[\log2x = \log \frac{ 5 }{ 3 }\]

now you can figure the rest out. I know you can do it!

Answer 4

you could also approach it \[\log6x = \log5\]

Answer 5

okay i got 0.737??

Answer 6

yes is 6x, thus
$$ \large {
log(2x)+log(3) = log(5) \implies log(2x \times 3) = log(5)\\
\implies log(6x) = log(5)\\
\text{now usig the log cancellation rule of }\\
a^{log_ax} = x\\
log(6x) = log(5) \implies log_{10}(6x) = log_{10}(5)\\
10^{log_{10}(6x)} = 10^{log_{10}(5)}\\
6x = 5
}
$$

Answer 7

you should get 5/6

Answer 8

ohh, i see what i did wrong! alright thank youuu

Answer 9

yw