This topic is taught in Secondary 3 after introduction of Indices Law.

In solving indices equation involving the same base, one of the common techniques is by **Substitution. **But** **before you can do substitution, you need to apply indices law to **'break down'** the equation. This process of breaking down is sometimes challenging for students. Knowing how to solve quadratic equation is also essential.

Sometimes, solving Indices Equation will also involve the concept of** taking lg** on both sides as well.

In the following example, you will Substitution and 'Breaking down' in action:

this is good

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You are such a great Mathematics. Iam impressed with your method of solving complex equations

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Yes you deserve a big up congratulation.Mathematics is no longer turf subject!

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how would you solve this if ^ means to the power of-

2^(x)3^(y)=6

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j hate Reply:

March 3rd, 2010 at 10:35 pm

with 2 unknowns you need 2 equations to solve for the unknowns

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how would you solve this if ^ means to the power of-

(2^2x+1)9^x=6^x

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wow this is really helpful, thank you soo much!

can someone help me with this? please

4^p-3 * 7^q-1 ?

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no root exist

Possible derivation:

d/dp(4^(p-3) 7^(q-1))

| Factor out constants:

= | 7^(q-1) (d/dp(4^(p-3)))

| Use the chain rule, d/dp(4^(p-3)) = ( d4^u)/( du) ( du)/( dp), where u = p-3 and ( d4^u)/( du) = 4^u log(4):

= | 7^(q-1) (4^(p-3) log(4) (d/dp(p-3)))

| Differentiate the sum term by term:

= | 4^(p-3) 7^(q-1) log(4) (d/dp(p)+d/dp(-3))

| The derivative of -3 is zero:

= | 4^(p-3) 7^(q-1) log(4) (d/dp(p)+0)

| The derivative of p is 1:

= | 1 4^(p-3) 7^(q-1) log(4)

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x5=125*squreroot of 2 pls solve d equation

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x^(x+1)+x^(2-x)-5^3-1=0

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write in the form of a power of m/n

?(ab-1)

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Can someone help me with this: show that (8x^2)^8-r (1/2x)=2^24-4r(X^16-3r)

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solve for x and y

x^x+y^y=31

x^y+y^x=17

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Praveen Reply:

March 21st, 2014 at 10:04 am

Sounds and looks complicated but actually not too bad once we see the trick. Maybe the qn should include the additional stipulation that x and y are integers.

Then, by simple trial and error, x and y are 2 and 3 (Either way).

Nice qn.

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This is so helpful:)

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I got lost at the quadratic equation it's not clear how you got (y-9)(y+1)=0

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Solve for:3^2y_6(3^y)=27

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Can sumone plz solve dis Indices for me.

Simplyfy -10a^2b^3[(-5a)^2/3b-1/3]

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.Cn sum1 help me wit dix plx

4/25^-1/2, multiply by 2^4 divid by (15/2^-2.

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pls solve X^2=16^X for me

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Ai Ling Reply:

February 20th, 2014 at 7:51 pm

May I know which grade you are?

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Praveen Reply:

March 21st, 2014 at 9:59 am

Interesting qn Rahim.

The answer is -0.5 I believe. You can either use graphing calculator or 'trial and error' method. I don't think you can do such questions using conventional logarithmic techniques.

If the qn were slightly modified, example x^2=3^x, and you are not allowed to use graphing calculator, then may need to rely on more advanced techniques such as Newton-Raphson method.

Comments and sharing are welcome.

Cheers

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Help on this

(81^2)*×(27^*)÷(9^*)=729

Where * represents the unknown

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Would like a help on

X^2= 16^x

Thanks

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