How do you solve #2^(x-3)=32#?

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Answer 1 · 2018-07-04T20:22:46

#color(blue)(x=8)#

#2^(x-3)=32#

Notice:

#32=2^5#

Therefore:

#2^(x-3)=2^5#

Since both bases are equal, then both exponents are equal:

#:.#

#x-3=5#

#x=8#

Answer 2 · 2018-07-05T02:29:57

#color(purple)(x = 8#

#2^(x - 3) = 32#

#2^(x - 3) = 2^5, " as " 32 = 2^5#

#(x - 3)cancel( log 2) =5 cancel(log 2), " taking log on both sides"#

#x - 3 = 5 " or " x = 8#

Answer 3 · 2018-07-05T03:10:48

#x=8#

When bases are equal, so are the exponents. We can rewrite #32# as #2^5# to get

#2^(x-3)=2^5#

We have the same base, so let's equate the exponents:

#x-3=5#

Adding #3# to both sides gives us

#x=8#

Hope this helps!