How do you solve for x #5^(log_5 *8) = 2x#?

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Answer 1 · 2015-10-02T22:24:59

#x=0.4#

To make things easy to write I'll call:

#log_(5)[0.8]=z#

So:

#5^z=2x#

Taking logs to base 5 of both sides#rArr#

#log_(5)5^z=log_(5)2x#

#zlog_(5)5=log_(5)2x#

Since #5^1=5:#

#log_(5)5=1#

So:

#zcancel(log_(5))5=log_(5)2x#

#z=log_(5)2x#

So:

#log_(5)[0.8]=log_(5)2x#

So:

#0.8=2x#

#x=0.4#