How do you solve #10 ^( 2x+1 )= 50#?

1 Answer
Mar 7, 2016

x = #( log 5 )/2#

Explanation:

log #a^m# = m log a.
log #10^(2x+1)# = (2x +1) log10 and log 10 =1.
log (ab) = log a + log b.
log 50 = log 5 + log 10 = log 5 + 1.
Logarithms (of any base ) of expressions on both sides are equal
The convenient choice here is common logarithm.
2 x = log 5..