In previous chapters we've learnt to simplify expressions involving indices (or powers) using index laws, but all the powers have so far been numbers. In exponentials, we are more interested about what happens if we raise the same base to different powers, so we'll see a lot of algebraic expressions in the powers. But don't worry, the same index laws still apply! To remind ourselves they are:
Be careful when using these rules as the bases have to be the same! For example, if we have an expression such as $3^x\times9^x$3x×9x we can not apply any of these rules as $3\ne9$3≠9. However if we think of $9$9 as two factors of $3$3 multiplied together we can proceed as follows:
$3^x\times9^x$3x×9x | $=$= | $3^x\times\left(3^2\right)^x$3x×(32)x |
$=$= | $3^x\times3^{2x}$3x×32x | |
$=$= | $3^{x+2x}$3x+2x | |
$=$= | $3^{3x}$33x |
Let's have a look at how to simplify (simplest index form) some more complicated examples!
Simplify the expression $\frac{6^{p-4q}\times36^{2q+p}}{6^{5p-3q}}$6p−4q×362q+p65p−3q
Think: First make sure all terms have the same base and be careful of order of operations when subtraction is concerned
Do:
$\frac{6^{p-4q}\times36^{2q+p}}{6^{5p-3q}}$6p−4q×362q+p65p−3q | $=$= | $\frac{6^{p-4q}\times\left(6^2\right)^{2q+p}}{6^{5p-3q}}$6p−4q×(62)2q+p65p−3q |
$=$= | $\frac{6^{p-4q}\times6^{2\left(2q+p\right)}}{6^{5p-3q}}$6p−4q×62(2q+p)65p−3q | |
$=$= | $\frac{6^{p-4q}\times6^{4q+2p}}{6^{5p-3q}}$6p−4q×64q+2p65p−3q | |
$=$= | $\frac{6^{p-4q+4q+2p}}{6^{5p-3q}}$6p−4q+4q+2p65p−3q | |
$=$= | $\frac{6^{3p}}{6^{5p-3q}}$63p65p−3q | |
$=$= | $6^{3p-\left(5p-3q\right)}$63p−(5p−3q) | |
$=$= | $6^{3p-5p+3q}$63p−5p+3q | |
$=$= | $6^{3q-2p}$63q−2p |
If $6^x=13$6x=13, find the value of $6^{x+1}$6x+1.
Use index laws to simplify the ratio $\left(0.3\right)^a:\left(0.3\right)^{a+2}$(0.3)a:(0.3)a+2 to a ratio of whole numbers.
Manipulate rational, exponential, and logarithmic algebraic expressions
Apply algebraic methods in solving problems