New Zealand
Level 7 - NCEA Level 2

# Logs and power functions reducible to quadratics

## Interactive practice questions

Consider the equation

$\left(2^x\right)^2-9\times2^x+8=0$(2x)29×2x+8=0

a

The equation can be reduced to a quadratic equation by using a certain substitution.

By filling in the gaps, determine the correct substitution that would reduce the equation to a quadratic.

Let $m=\left(\editable{}\right)^{\editable{}}$m=()

b

Solve the equation for $x$x by using the substitution $m=2^x$m=2x.

Easy
Approx 3 minutes

Solve the following equation for $x$x by substituting in $m=4^x$m=4x.

$4^{2x}-65\times4^x+64=0$42x65×4x+64=0

Use a comma to separate multiple solutions.

Solve for $x$x:

$2^{2x}-12\times2^x+32=0$22x12×2x+32=0

Let $p=2^x$p=2x.

Solve for $x$x:

$4\times2^{2x}-34\times2^x+16=0$4×22x34×2x+16=0

### Outcomes

#### M7-6

Manipulate rational, exponential, and logarithmic algebraic expressions

#### 91261

Apply algebraic methods in solving problems