[SOLVED] Checking Inequalities

• Apr 12th 2010, 11:22 AM
anindyog
[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 AM
e^(i*pi)
Quote:

Originally Posted by anindyog
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: $a^x > 0$

b) First things first: $x \neq 6$

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 $x > 6$ and we do not need to worry about the direction of the inequality changing

This gives $4 > 5x-30$ and you can solve for x in the normal way.

$x < \frac{34}{5}$
• Apr 12th 2010, 12:12 PM
Soroban
Hello, anindyog!

Quote:

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

Multiply by 4: . $4^x + 2 \:=\:3\!\cdot\!2^x \quad\Rightarrow\quad (2^2)^x + 2 \:=\:3\!\cdot\!2^x \quad\Rightarrow\quad 2^{2x} - 3\!\cdot\!2^x + 2 \:=\:0$

Factor: . $\left(2^x - 1\right)\left(2^x - 2\right) \:=\:0$

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

• Apr 12th 2010, 02:14 PM
anindyog
Thanks alot! However how do you get to 4^x from 4^(x-1) after multiplying it by 4?
• Apr 12th 2010, 10:53 PM
mr fantastic
Quote:

Originally Posted by anindyog
Thanks alot! However how do you get to 4^x from 4^(x-1) after multiplying it by 4?

You're expected to be familiar with index laws such as $a^n \cdot a^m = a^{n + m}$.