Hello Peyton Sawyer Originally Posted by
Peyton Sawyer solve following equations...
Here are some hints. See if you can continue.
$\displaystyle \ln(3x)=4$
$\displaystyle \Rightarrow 3x = e^{\text ?}$
$\displaystyle e^{0.005x}=2$
$\displaystyle \Rightarrow 0.005x = \ln(\text ?)$
$\displaystyle 10^{x^2}=10^{3x-2}$
$\displaystyle \Rightarrow x^2=3x-2$
I assume this means:$\displaystyle 6^{x+1}=4^{2x-1}$
$\displaystyle \Rightarrow (x+1)\log 6 = (2x-1)\log 4$ (where the log is to any base you like)
Collect like terms to solve for $\displaystyle x$ in the usual way.
Doesn't make sense.
Does this mean:$\displaystyle \log_3 (1-x)=1$?
If so:$\displaystyle 1-x = 3^1$
$\displaystyle \Rightarrow \text ?$
$\displaystyle \log(x+1)(x-1)= \log 3$
$\displaystyle \Rightarrow \text ?$
Grandad