HELP!! Logarithmics solving for x.?

Question

a.) 7.2^x-65=0
b.) log 2 (x+1)=3

Answer 1

you should know this logarithm properties:

log a^c = c log a

log(base 10) a = b
a = 10^b
so,
a)7.2^x-65=0
7.2^x = 65 take log for both side
log (7.2^x) =log 65
x log 7.2 = log 65
x = log 65 /(log7.2) =2.115

b.) log 2 (x+1)=3 (if the base10)
2(x+1)=10^3
2x+2 =1000
2x=1000-2
x=998/2=499

Answer 2

a) Here is a similar one

5^x - 4 = 0

5^x = 4

log(5^x) = log4

xlog5 = log4

x = log5/log4

Now try yours

b) Try to remember these basic rule

2^3 = 8 is identical to 3 = log 8 base2

log 32 base 2 = 5 is identical to 2^5 = 32





so log 4(x+ 3) = 2 is identical to 10^2 = 4(x + 3)

100 = 4x + 12

4x = 88

x = 22

Now try your 2 questions. The answers should be 499 and 2.11

Answer 3

a) 7*2^x=65
so 2^x=65/7 and taking log
x log2 = log(65/7) so x= log(65/7)/log2 =3.215
b) if you wrote log 2(x-1) =3
2(x+1) = 10^3 so x+1 = 500 and x= 499
If meant log(base2)(x+1) = 3
x+1= 2^3 =8 and x=7

Answer 4

Divide by 3^[x+a million] a million + a million/3 = 36/3^[x+a million] 4/3 = 36/3^[x+a million] 3^[x+a million] = 27 yet 3^3 = 27 So [x + a million] = 3 x = 2 i think of this grew to become into the physique of suggestions you have been attempting out of your 3^-a million you do no longer ought to bypass into logs

Answer 5

for part b the answer is 7. ok what you do is look at it this way 2^3=x plus 1 so 8= x plus 1 then subtract 1 from both sides and get 7

Answer 6

7.2^x = 65
take log with base 7.2 of both sides
log (base 7.2) 7.2^x = log (base 7.2) 65
x = log 65/log 7.2
x = 2.11