More logarithms?

Question

Solve for X: 16=2^7x-5

Can someone answer this for me please? Im doing some Algebra homework on Logarithms and i quite dont understand well what im supposed to be doing.

Answer 1

i think you mean 16= 2^(7x-5)
2^4 = 2^(7x-5)
4 = 7x - 5
x = 9/7

if you meant 16 = 2^(7x) - 5
2^(7x) = 21
7x ln(2) = ln (21)
x = ln(21) / ( 7 ln 2)

if you meant 16 = (2^7)x -5
16 = 128x - 5
x = 21/128

Answer 2

I suppose you wrote
16= 2^(7x-5)
in this case taking log in base 2 as 16=2^4 and log 2 =1

4 = 7x-5 and x= 9/7

Answer 3

I assume you meant:
16 = 2^(7x - 5)

The parentheses are necessary to show that the subtraction happens on the exponent.

16 = 2^(7x - 5)

Take the log(base 2) of both sides...
log(base 2) 16 = log(base 2) [2^(7x - 5)]

Now, notice that log(base 2) [2^x] = x...
log(base 2) 16 = 7x - 5

Simplify the log...
4 = 7x - 5

Now solve for x as you would always do...
7x = 9
x = 7/9

Answer 4

OK...
16=2^(7x) - 5
21 = 2^(7x)
Log_base2(21) = 7x
x = (log-base2(21))/7

Get your calculator.