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

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- Apr 12th 2010, 11:22 AManindyog[SOLVED] Checking Inequalities
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 - Apr 12th 2010, 11:31 AMe^(i*pi)
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.

$\displaystyle x < \frac{34}{5}$ - Apr 12th 2010, 12:12 PMSoroban
Hello, anindyog!

Quote:

$\displaystyle (a)\;\; 4^{x - 1} + \tfrac{1}{2}\;=\;\tfrac{3}{4}\!\cdot\!2^x$

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: . $\displaystyle \begin{Bmatrix}\;2^x-1\:=\:0 & \Rightarrow & 2^x \:=\:1 & \Rightarrow & x \:=\:0\; \\ \\[-3mm]2^x-2 \:=\:0 & \Rightarrow & 2^x \:=\:2 & \Rightarrow & x \:=\:1 \end{Bmatrix}$

- Apr 12th 2010, 02:14 PManindyog
Thanks alot! However how do you get to 4^x from 4^(x-1) after multiplying it by 4?

- Apr 12th 2010, 10:53 PMmr fantastic