I need help!
how to solve
log10x^7.. expanding with the rules..
9e^5x=90
and 2^6-x=10
I would greatly appreciate the help! I know you guys were so urgent to help last time I needed it!
TAHNKSS!!
Please clarify:
do you mean $\displaystyle \log_{10}x^7$
and
$\displaystyle 2^{6 - x} = 10$?
For the other question:
$\displaystyle 9e^{5x} = 90$
$\displaystyle \Rightarrow e^{5x} = 10$
$\displaystyle \Rightarrow \ln e^{5x} = \ln 10$
$\displaystyle \Rightarrow 5x \ln e = \ln 10$ .................since $\displaystyle log_a (x^n) = n \log_a x$
$\displaystyle \Rightarrow 5x = \ln 10$ ....................since $\displaystyle \log_a a = 1$ so in particular, $\displaystyle \ln e = 1$
$\displaystyle \Rightarrow x = \frac {\ln 10}5$
$\displaystyle \log 10x^7 = \log 10 + \log x^7$ ..............since $\displaystyle \log_a xy = \log_a x + \log_a y$
$\displaystyle = \log 10 + 7 \log x$ .................since $\displaystyle \log_a (x^n) = n \log_a x$
$\displaystyle = 1 + 7 \log x$ .......................if the log is the base 10 logarithm. since $\displaystyle \log_a a = 1$
$\displaystyle 2^{6 - x} = 10$
$\displaystyle \Rightarrow \ln 2^{6 - x} = \ln 10$
$\displaystyle \Rightarrow (6 - x) \ln 2 = \ln 10$
$\displaystyle \Rightarrow 6 - x = \frac {\ln 10}{\ln 2}$
$\displaystyle \Rightarrow x = 6 - \frac {\ln 10}{\ln 2}$
Okay, I got that one! Thank you..
Now... haha sorry..
log3 (x+1) + log3 (x+3)=1
the answer is there is no solution, 0, because you can't have a take a negative log.. i tried to solve it and i combined them and did log3 (x^2+4x+3)=1 but i have no idea where to go from there