How do you solve #2^(x+3)=5^(3x-1)#?

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Answer 1 · 2016-09-19T02:53:46

#x = ln40/ln62.5 ~= 0.892#

Take the natural logarithm of both sides:

#ln(2^(x + 3)) = ln(5^(3x- 1))#

Apply the rule #ln(a^n) = nlna#:

#(x + 3)ln2 = (3x - 1)ln5#

#xln2 + 3ln2 = 3xln5 - ln5#

Isolate the x's to one side:

#3ln2 + ln5 = 3xln5 - xln2#

#3ln2 + ln5= x(3ln5 - ln2)#

Re-condense using the rule #nlna = ln(a^n)#.

#ln8 + ln5 = x(ln125 - ln2)#

Simplify both sides using the rule #lna + lnb = ln(a xx b)# and #lna - lnb = ln(a/b)#

#ln(8 xx 5) = x(ln(125/2))#

#ln(40) = x(ln(125/2))#

#x = ln40/ln62.5#

If you want an approximation, you will get #x ~=0.892#.

Hopefully this helps!