How do you solve #62^(x+3) <= 7^(2x+1)#?

take the common logarithm of both sides and you will get:

#log62^(x+3)<=log7^(2x+1)#

using the power rule for logs, the exponents become factors or multipliers and the equation reduces to:

#(x+3)log62<=(2x+1)log7#
by expanding , we will obtain a simple linear equation with 1 variable (x):

xlog62 + 3log62 #<=# 2xlog7 + 1log7

collect all your x terms on one side of the equation and the others on the opposite side:

xlog62 - 2xlog7 #<=# log7 - 3log62

factoring the common factor of 'x':

x(log62 - 2log7) #<=# log7 - 3log62

and the result is:

x #<=# #(log7-3log62)/(log62-2log7)#

You could use any form of log for this. I chose #log_e#

Taking logs of both sides

#" "(x+3)ln(62)color(red)(<=)(2x+1)ln(7)#

#" "=>(ln(62))/(ln(7))color(red)(<=)(2x+1)/(x+3)#

Dividing the right hand side gives

#" "(ln(62))/(ln(7))color(red)(<=)2 -5/(x+3)#

#" "(ln(62))/(ln(7))-2color(red)(<=) -5/(x+3)#

Multiply by (-1)

#" "2-(ln(62))/(ln(7))color(red)(>=) +5/(x+3)#

#" "(x+3)(2-(ln(62))/(ln(7)))color(red)(>=) +5#

#" "x+3 color(red)(>=) 5 xx (ln(7))/(2ln(7)-ln(62))#

#" "xcolor(red)(>=) (5ln(7))/(2ln(7)-ln(62)) -3#

#" "xcolor(red)(>=) (ln(7^5))/(ln(7^2/62))-3#