# If a^(x+3) = b^(-1), then x =

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When a>0 and b>0

`a^(x+3)=b^-1`

`(x+3)loga=-logb`

`(x+3)=-logb/loga`

or, `x=-3-logb/loga`

When a<0 the problem has no slution at all.

when a=0 and b=0 , the problem has no slution. Therefore the only possibility to solve the above equation is that a>0 and b>0.

To solve for this, you simplify first the right side by applying this property of negative exponent:

`n^(-1) = 1/n`

`` Rewritng the problem, you have:

`a^(x+3) = 1/b`

Take the logarithm of both.

`loga^(x+3) = log(1/b)`

Apply `logM^n = nlogM` on the left side and `log(M/N) = logM - logN` on the right side.

`(x+3)loga = log1 - logb`

recall that log 1 = 0.

`(x+3) log a = 0 - logb`

The, divide both sides by log a so you can isolate the terms with x on one side.

`(x+3)*loga/(loga) = -logb/(loga)`

`x+3 = -logb/(loga)`

Subtract 3 on both sides so x can be isolated on the left side.

`x+3-3 = -logb/(loga) - 3`

Therefore, `x = -logb/(loga) - 3`

`a^(x+3)=b^(-1)`

We want to isolate x, but there is no simple step to do that just yet since it is in the exponent form. So we must first take the log of both sides and apply the property: `log a^x = xloga`

``

`=> log a^(X+3) = log b^(-1)`

`=> (x+3)loga = -1logb`

Then we can employ simple algebra to isolate x.

`=> x+3 = -logb/loga`

**`=> x = (-logb/loga)-3`**