How do you solve $8^{2 x} = 8^{x + 7}$ $8^(2x) = 8^(x+7)$ ?

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Answer 1 · 2016-04-05T03:48:04

$x = 7$ $x=7$

Note that since the bases of both exponential functions are equal, their exponents must also be equal.

This gives us, once we set their exponents equal to one another:

$2 x = x + 7 \Rightarrow x = 7$ $2x=x+7" "=>" "x=7$

If we want to formalize this, take the logarithm with base $8$ $8$ of both sides to undo the exponential functions.

${log}_{8} (8^{2 x}) = {log}_{8} (8^{x + 7}) \Rightarrow 2 x = x + 7$ $log_8(8^(2x))=log_8(8^(x+7))" "=>" "2x=x+7$

Which is the equation we saw previously.

We can generalize this idea of exponential functions having the same bases as saying that if:

$a^{b} = a^{c} \Rightarrow b = c$ $a^b=a^c" "=>" "b=c$

How do you solve 82x=8x+78^(2x) = 8^(x+7)?