How do you solve $12^{2 x - 3} = 16$ $12^(2x-3)=16$ ?

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How do you solve $12^{2 x - 3} = 16$ $12^(2x-3)=16$ ?

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Answer 1 · 2015-07-04T00:34:50

This suggests that you can use ${log}_{12} (x)$ $log_12(x)$ , where $x$ $x$ is either $12^{2 x - 3}$ $12^(2x-3)$ or $16$ $16$ . (By default, $log x = {log}_{10} x$ $log x = log_10 x$ .)

${log}_{12} (12^{2 x - 3}) = {log}_{12} (16)$ $log_12(12^(2x-3)) = log_12(16)$

Now if you use the change of base law, you get:

$\frac{log (12^{2 x - 3})}{log 12} = \frac{log 16}{log 12}$ $(log (12^(2x-3)))/(log12) = (log 16)/(log 12)$

Since $\frac{ln x}{log x} = constant$ $(lnx)/(logx) = "constant"$ (try it on any $x$ $x$ ; it is $\approx 2.303$ $~~2.303$ ), we can change this to:

$\frac{ln (12^{2 x - 3})}{ln 12} = \frac{ln 16}{ln 12}$ $(ln (12^(2x-3)))/(ln12) = (ln 16)/(ln 12)$

Using exponential rules of logs:

$(2x-3)cancel(((ln 12)/(ln12))) = (ln 16)/(ln 12)$

$2x - 3 = (ln 16)/(ln 12)$

$x = color(blue)(((ln 16)/(ln 12) + 3)/2 ~~ 2.05789)$

If you plug this back in, it indeed works.

$12^((2(~~2.05789) - 3)) = 16$

$12^((~~1.11578)) = 16$

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How do you solve $12^{2 x - 3} = 16$ $12^(2x-3)=16$ ?

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