How do you solve #2^x*5=10^x#?

2 Answers
Jun 30, 2016

If #A = B#, then #logA = logB#:

#log(2^x xx 5) = log(10^x)#

Use the rule #log_a(n/m) = log_an - log_am# to simplify.

#log2^x + log5 = log10^x#

Use the rule #loga^n = nloga# to simplify further.

#xlog2 + log5 = xlog10#

#log5 = xlog10 - xlog2#

Factor out the #x#.

#log5 = x(log10 - log2)#

Put back the logs on the left side into quotient form to simplify.

#log5 = x(log(10/2))#

#log5 = x(log5)#

Isolate x.

#log5/log5 = x#

#1 = x#

Checking in the original equation, we find that this solution works.

Hopefully this helps!

Jun 30, 2016

Answer:

#x=1# (using alternate solution method)

Explanation:

If #2^x*5=10^x#

#rArr 2^x*5 =(2*5)^x#

#rArr 2^x*5 = 2^x*5^x#

#rArr 5=5^x#

#rArr x=1#