How do you solve #1-2e^(2x)=-19#?

3 Answers
May 1, 2018

Answer:

# x = ln \sqrt{10} #

Explanation:

# 1 - 2 e^{2x } = -19 #

# -2 e^{2x} = -19 -1 = -20 #

# e^{2x} = -20/(-2) = 10 #

# ln e^{2x} = ln 10 #

# 2x = ln 10 #

# x = {ln 10}/2 = ln \sqrt{10} #

Check:

# 1 - 2 e^{2x } #

#= 1 - 2 e^{2 (ln sqrt{10} ) } #

#= 1 - 2 e^{ln 10} #

# = 1 - 2(10) #

# = -19 quad sqrt #

Answer:

the value is #~~1.151#

Explanation:

given #1-2e^(2x)=-19rArr-2e^(2x)=-20rArre^(2x)=10#
in general we have #e^m=krArr log_ek=m#
which means we have #log_e10=2x# and #log_e10~~2.302#
we have #2x=2.302rArrx~~1.151#

May 1, 2018

Answer:

#x = (ln10)/2#
#~~1.1512925465#

Explanation:

Subtract 1 on both sides.
#-2e^(2x) = -20#
Divide by -2.
#e^(2x) = 10#
Taking the logarithm of both sides, we have:
#ln(e^(2x)) = ln10#
Using the power rule of logarithms,
#2xln(e) = ln 10#

#lne = 1# So, we have:

#2x = ln 10#
#x = (ln10)/2#