How do you solve for x in $6 = 15^{1 - x}$ $6 = 15^(1-x)$ ?

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Answer 1 · 2016-05-02T20:32:29

Use logs to calculate: $x = 0.338$ $x = 0.338$

The variable is in the index and 6 is definitely not a power of 15, therefore logs are called for...

Log both sides: $log 6 = {log 15}^{1 - x}$ $log 6 = log 15^(1-x)$

Power law: #log 6 = (1-x) log 15

isolate (1-x): $\frac{log 6}{log 15} = 1 - x$ $(log6)/(log15) = 1-x$

With a calculator: $1 - x = 0.6616$ $1-x = 0.6616$

                             #0.338 = x#

How do you solve for x in 6=151−x6 = 15^(1-x)?