How do you solve $7^{2 x - 3} - 4 = 14$ $7^(2x-3) - 4 = 14$ ?

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Answer 1 · 2016-04-06T03:50:49

You must work with the fact that $a^{n} = m \to {log a}^{n} = log m$ $a^n = m -> loga^n = logm$

$7^{2 x - 3} = 18$ $7^(2x - 3) = 18$

${log 7}^{2 x - 3} = log 18$ $log7^(2x - 3) = log18$

Simplify using the rule ${log a}^{n} = n log a$ $loga^n = nloga$

$(2 x - 3) log 7 = log 18$ $(2x - 3)log7 = log18$

$2 x log 7 - 3 log 7 = log 18$ $2xlog7 - 3log7 = log18$

$2 x log 7 = log 18 + 3 log 7$ $2xlog7 = log18 + 3log7$

Use the rule ${log}_{a} n + {log}_{a} m = {log}_{a} (n \times m)$ $log_an + log_am = log_a(n xx m)$

$x (2 log 7) = log (18 \times 343)$ $x(2log7) = log(18 xx 343)$

$x = \frac{log 6174}{log 49}$ $x = log6174/log49$

$x = 2.24$ $x = 2.24$

Make sure to ask your teacher whether or not they require your answer in logarithmic (exact) form.

Hopefully this helps!

How do you solve 72x−3−4=147^(2x-3) - 4 = 14?