How do you solve $4^{2 x} = 6$ $4 ^(2x) = 6$ ?

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How do you solve $4^{2 x} = 6$ $4 ^(2x) = 6$ ?

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Answer 1 · 2016-03-28T16:20:07

You must use logarithms to solve.

Explanation:

If $a^{n} = b$ $a^n = b$ , then ${log a}^{n} = log b$ $loga^n = logb$ .

${log 4}^{2 x} = log 6$ $log4^(2x) = log6$

Use the log rule ${log a}^{n} = n log a$ $loga^n = nloga$ to simplify further.

$2 x log 4 = log 6$ $2xlog4 = log6$

$x ({log 4}^{2}) = log 6$ $x(log4^2) = log6$

$x = \frac{log 6}{log 16}$ $x = (log6)/(log16)$

$x = 0.65$ $x = 0.65$

Make sure to ask your teacher if they want your answer in exact form ( $x = \frac{log 6}{log 16}$ $x = (log6)/(log16)$ ) or rounded off to an x number of decimal points. I know that my teacher asks for two decimals after the point ( $x = 0.65$ $x = 0.65$ ), but some ask for 1, 3 or 0.

Practice exercises:

Solve for x:

$a) 2^{3 x - 2} = 17$ $a) 2^(3x - 2) = 17$

$b) 3^{4 x + 1} = 6^{3 x - 3}$ $b) 3^(4x + 1) = 6^(3x - 3)$

Challenge problem:

Solve for x in $2^{2 x - 3} + 5 = 3^{3 x}$ $2^(2x - 3) + 5 = 3^(3x)$

Hint: think of the log addition rule.

Good luck!

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How do you solve $4^{2 x} = 6$ $4 ^(2x) = 6$ ?

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Explanation:

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