Solve exponential equation with exponent variables.

Question

Ok, the question is:

Solve the following for x:

$2^x4^{x-1}=70$

I have just asked Wolfram Alpha, of course though, it supplies an answer without revealing its working.

I started to try the product of the bases and the sum of the exponents, giving me:

$8^{2x-1}$ ?

But Alpha says I'm barking up the wrong tree?

With thanks in advance!

Answer 1

The issue is that you cannot add exponents with different bases. You cannot multiply bases with different exponents. The trick is this: $4=2^2$. Then we can rewrite this as $$2^x4^{x-1}=2^x(2^2)^{(x-1)}=2^x2^{2(x-1)}=2^x2^{2x-2}.$$ Now we can add exponents: $$2^{3x-2}=70.$$ Try this.

Answer 2

eqn = 2^x 4^(x - 1) == 70;

For real solution using Mathematica

soln = Solve[eqn, x, Reals][[1]]

(*  {x -> (3*Log[2] + Log[5] + Log[7])/
       (3*Log[2])}  *)

Verifying that soln satisfies eqn

eqn /. soln

(*  True  *)

Wolfram|Alpha has a button for step-by-step solution. Using Wolfram|Alpha from within Mathematica

WolframAlpha["solve 2^x 4^(x\[Minus]1)==70 for x", {{"RealSolution", 2}, 
  "Content"}, PodStates -> {"RealSolution__Step-by-step solution"}]

This is equivalent to earlier result

(x /. soln) == 2/3 + Log[70]/(3 Log[2])

(*  True  *)

Answer 3

You are probably expected to notice that $4=2^2$ and write your expression as $2^{3x-2}$. However, this is not necessary in this case.

Let $\log$ be the logarithm to any base. Recall that if $a$ and $b$ are positive then $\log(ab)=\log a+\log b$, and that $\log(a^k)=k\log a$.

So from $2^x4^{x-1}=70$ we obtain, by taking the logs of both sides, $$x\log 2+(x-1)\log 4=\log(70).$$ It follows that $$x\log 2+x\log 4=\log(70)+\log 4,$$ and therefore $$x=\frac{\log{70}+\log 4}{\log 2+\log 4}.$$ One can get a simpler expression by noting that $4=2^2$ and therefore $\log 4=2\log 2$.