Sorry I'm quite bad at maths, but wanted to find out
a) 4^(x - 1) + 0.5 = (3/4).2^x
and
b) 4/(x - 6) > 5
thanks
The first one is an equation, not an inequality. Remember that for real x and positive a: $\displaystyle a^x > 0$
b) First things first: $\displaystyle x \neq 6$
Since the right hand side is positive then it follows that the left hand side must also be positive, this sets a lower bound of $\displaystyle x > 6$ and we do not need to worry about the direction of the inequality changing
This gives $\displaystyle 4 > 5x-30$ and you can solve for x in the normal way.
$\displaystyle x < \frac{34}{5}$
Hello, anindyog!
$\displaystyle (a)\;\; 4^{x - 1} + \tfrac{1}{2}\;=\;\tfrac{3}{4}\!\cdot\!2^x$
Multiply by 4: .$\displaystyle 4^x + 2 \:=\:3\!\cdot\!2^x \quad\Rightarrow\quad (2^2)^x + 2 \:=\:3\!\cdot\!2^x \quad\Rightarrow\quad 2^{2x} - 3\!\cdot\!2^x + 2 \:=\:0 $
Factor: .$\displaystyle \left(2^x - 1\right)\left(2^x - 2\right) \:=\:0$
Therefore: . $\displaystyle \begin{Bmatrix}\;2^x-1\:=\:0 & \Rightarrow & 2^x \:=\:1 & \Rightarrow & x \:=\:0\; \\ \\[-3mm]2^x-2 \:=\:0 & \Rightarrow & 2^x \:=\:2 & \Rightarrow & x \:=\:1 \end{Bmatrix}$