How do you solve $4^{2 x + 3} = 1$ $4^(2x+3) = 1$ ?

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Answer 1 · 2016-04-02T01:25:05

For this problem, we must use the rule $a^{n} = b^{m} \to {log a}^{n} = {log b}^{m}$ $a^n = b^m -> loga^n = logb^m$

$4^{2 x + 3} = 1$ $4^(2x + 3) = 1$

${log 4}^{2 x + 3} = log 1$ $log4^(2x + 3) = log1$

Use the property ${log n}^{a} = a log n$ $logn^a = alogn$

$(2 x + 3) log 4 = 0$ $(2x + 3)log4 = 0$

$2 x + 3 = \frac{0}{log 4}$ $2x + 3 = 0/log4$

$2 x + 3 = 0$ $2x + 3 = 0$

$2 x = - 3$ $2x = -3$

$x = - \frac{3}{2}$ $x = -3/2$

Checking we find that the solution works, since $a^{0} = 1 \to 4^{0} = 1$ $a^0 = 1 -> 4^0 = 1$

Practice exercises:

Solve for x. Express your answer in terms of $log x$ $logx$ (exact values)

a) $2^{3 x + 1} = 5$ $2^(3x + 1) = 5$

b) $3^{2 x - 1} = 4^{3 x - 7}$ $3^(2x - 1) = 4^(3x - 7)$

Challenge Problem

Solve $2^{3 x} \times 5 = 3^{4 x - 1}$ $2^(3x) xx 5 = 3^(4x - 1)$

Good luck!

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