# 8^x+1 = (1/2)^x+5

#### beautifulconfusion

##### New member
I'm having trouble with this problem from an assignment. Someone on a different message board said: this problem it is not possible. literally not possible you must have written it wrong.

It sure looks possible to me, I just can't figure it out :|

#### tkhunny

##### Moderator
Staff member
Please try harder with your notation. I'm pretty sure you mean 8^(x+1) = (1/2)^(x+5). Extra parentheses clarify meaning.

Once this is done, it is easy enough IF you notice that both bases are integer powers of 2.

8 = 2^3

1/2 = 2^(-1)

This leads to 3(x+1) = (-1)(x+5)

Now what?

#### beautifulconfusion

##### New member
Everyone seems to think I am leaving out ()'s but I'm not. This is how it is on the assignment. I'm very careful.
Sorry, I entered the wrong image. Here it is.

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#### stapel

##### Super Moderator
Staff member
beautifulconfusion said:
Everyone seems to think I am leaving out ()'s but I'm not.... Here it is.
So the first reply was correct: you were omitting parentheses.

Standard web-safe formatting, along with the necessity for using grouping symbols, was amply explained and illustrated in the formatting articles in the links in the "Read Before Posting" thread that you read before posting.

As those articles explained, your posting of "8^x + 1 = (1/2)^x + 5" meant "8[sup:170xuql4]x[/sup:170xuql4] + 1 = (1/2)[sup:170xuql4]x[/sup:170xuql4] + 5". This is obviously quite different from what you've posted in the image; namely, "8^(x + 1) = (1/2)^(x + 5)" which means "8[sup:170xuql4]x+1[/sup:170xuql4] = (1/2)[sup:170xuql4]x+5[/sup:170xuql4]". :wink:

Eliz.

#### Subhotosh Khan

##### Super Moderator
Staff member
Now tell us:

$$\displaystyle 2^{?}\, = \, 8$$

and

$$\displaystyle 2^{?}\, = \, \frac{1}{2}$$