ðŸ‘‰This post extends our series on indices ðŸ‘ˆ

If an exponential equation has the same base on each side of the equation, then the exponent is equal.

Using this fact, we can solve the following

**Example 1**

Solve for $x$ in the equation

$7^{5x+8}=7^{x^2+3x}$

**Solution:**

The base($7$) is the same on both sides, hence the exponents are equal.

$5x+8=x^2+3x$

Re-arranging the equation

$0=x^2+3x-5x-8$

$0=x^2-2x-8$

$x^2-2x-8=0$

$x^2-4x+2x-8=0$

$x(x-4)+2(x-4)=0$

$(x+2)(x-4)=0$

$x=-2$ or $x=4$

**Example 2**

Sole for $x$ in the equation

$4^x-2^{x+2}=32$

**Solution:**

Simplifying the indicial equation

$2^{2(x)}-2^x\times2^2=32$

$2^{x(2)}-2^x(4)=32$

Let $2^x=t$

$t^2-t(4)=32$

$t^2-4t-32=0$

By factorization

$t^2+4t-8t-32=0$

$t(t+4)-8(t+4)=0$

$(t-8)(t+4)=0$

$t=8$ or $-4$

$t$ can't be negative, hence

$t=8$, but $t≠-4$

Inserting the value of $t$ in $2^x=t$

$2^x=8$

$2^x=2^3$

The base is the same, therefore,

$x=3$

**Example 3**

Find the value of x in this equation

$5^{2x-3}-126(5^{x-3})+1=0$

**Solution**

$5^{2x-3}-126(5^{x-3})+1=0$

will translate to

$\frac{5^{2x}}{5^3}-\frac{126(5^x)}{5^3}+1=0$

$\frac{5^{x(2)}}{125}-\frac{126(5^x)}{125}+1=0$

Let $5^x=t$

$\frac{t^2}{125}-\frac{126t}{125}+1=0$

Taking the L.C.M

$\frac{t^2-126t+125}{125}=0$

Cross multiply

$t^2-126t+125=0\times125$

$t^2-126t+125=0$

Solving quadratically

$t^2-125t-1t+125=0$

$t(t-125)-1(t-125)=0$

$(t-1)(t-125)=0$

$t=1$ or $125$

Substituting the value of t in $5^x=t$

When t=1

$5^x=1$

$5^x=5^0$

$x=0$

When t=125

$5^x=125$

$5^x=5^3$

$x=3$

Therefore $x=3$ and $0$

__Related post__**Example 4**

Solve for x if $2^{2x+1}+2^x-1=0$

**Solution:**

$2^{2x}\times2+2^x-1=0$

$2^{x(2)}\times2+2^x-1=0$

Let $2^x=t$

$t^2(2)+t-1=0$

$2t^2+t-1=0$

By factorization

$2t^2+2t-t-1=0$

$2t(t+1)-1(t+1)=0$

$(2t-1)(t+1)=0$

$2t-1=0$ or $t+1=0$

$2t=1$ or $t=-1$

$\frac{2t}{2}=\frac{1}{2}$ or $t=-1$

$t=\frac{1}{2}$ or $t=-1$

$t$ cannot be negative, therefore,

$t=\frac{1}{2}$ but $t≠-1$

Inserting the value of t in $2^x=t$

$2^x=\frac{1}{2}$

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

$x=-1$

That will be all for now. In my next post, we shall continue our series on exponents when we look at exponential simultaneous equations.

In the meantime, if you have got questions, remember to ask our Facebook community. Likewise, you can also ask our telegram community.

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